Chapter 16: Electric Charge And Electric Field

1. A neutral atom always has

More neutrons than protons.

More protons than electrons.

The same number of neutrons as protons.

The same number of protons as electrons.

A neutral atom is one that has an equal number of protons and electrons. Protons have a positive charge, while electrons have a negative charge. The charges of protons and electrons cancel each other out, resulting in a neutral overall charge for the atom. Neutrons, on the other hand, have no charge and do not affect the overall charge of the atom. Therefore, a neutral atom will always have the same number of protons as electrons.

Explanation

A neutral atom is one that has an equal number of protons and electrons. Protons have a positive charge, while electrons have a negative charge. The charges of protons and electrons cancel each other out, resulting in a neutral overall charge for the atom. Neutrons, on the other hand, have no charge and do not affect the overall charge of the atom. Therefore, a neutral atom will always have the same number of protons as electrons.

2. Materials in which the electrons are bound very loosely to the nuclei and can move about freely within the material are referred to as

Insulators.

Conductors.

Semiconductors.

Superconductors.

Conductors are materials in which electrons are bound very loosely to the nuclei and can move about freely within the material. This allows for the easy flow of electric current through the material. Insulators, on the other hand, have tightly bound electrons that do not move easily, while semiconductors have properties in between conductors and insulators. Superconductors are materials that exhibit zero electrical resistance at very low temperatures.

Explanation

Conductors are materials in which electrons are bound very loosely to the nuclei and can move about freely within the material. This allows for the easy flow of electric current through the material. Insulators, on the other hand, have tightly bound electrons that do not move easily, while semiconductors have properties in between conductors and insulators. Superconductors are materials that exhibit zero electrical resistance at very low temperatures.

3. An atom has more electrons than protons. The atom is

A positive ion.

A negative ion.

A superconductor.

Impossible.

An atom having more electrons than protons indicates that it has gained extra electrons, resulting in a negative charge. This excess of negatively charged electrons makes the atom a negative ion.

Explanation

An atom having more electrons than protons indicates that it has gained extra electrons, resulting in a negative charge. This excess of negatively charged electrons makes the atom a negative ion.

4. Two charged objects attract each other with a certain force. If the charges on both objects are doubled with no change in separation, the force between them

Quadruples.

Doubles.

Halves.

Increases, but we can't say how much without knowing the distance between them.

When the charges on both objects are doubled, the force between them is directly proportional to the product of the charges. Since both charges are doubled, the product of the charges is quadrupled. Therefore, the force between them also quadruples.

Explanation

When the charges on both objects are doubled, the force between them is directly proportional to the product of the charges. Since both charges are doubled, the product of the charges is quadrupled. Therefore, the force between them also quadruples.

5. The model of the atom shows a

Neutrally charged nucleus surrounded by both protons and electrons.

Nucleus consisting of both protons and neutrons, surrounded by a cloud of electrons.

Nucleus consisting of both electrons and neutrons, surrounded by a cloud of protons.

Nucleus consisting of both protons and electrons, surrounded by a cloud of neutrons.

The correct answer is that the model of the atom shows a nucleus consisting of both protons and neutrons, surrounded by a cloud of electrons. This is based on the current understanding of atomic structure, where the nucleus contains the positively charged protons and neutral neutrons, while the negatively charged electrons orbit around the nucleus in energy levels or electron clouds.

Explanation

The correct answer is that the model of the atom shows a nucleus consisting of both protons and neutrons, surrounded by a cloud of electrons. This is based on the current understanding of atomic structure, where the nucleus contains the positively charged protons and neutral neutrons, while the negatively charged electrons orbit around the nucleus in energy levels or electron clouds.

6. Two point charges, separated by 1.5 cm, have charge values of 2.0 and -4.0 μC, respectively. What is the magnitude of the electric force between them?

400 N

360 N

320 N

160 N

The magnitude of the electric force between two point charges can be calculated using Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the charges are 2.0 and -4.0 μC, and the distance between them is 1.5 cm. Plugging these values into the equation, we can calculate the magnitude of the electric force to be 320 N.

Explanation

The magnitude of the electric force between two point charges can be calculated using Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the charges are 2.0 and -4.0 μC, and the distance between them is 1.5 cm. Plugging these values into the equation, we can calculate the magnitude of the electric force to be 320 N.

7. A proton carries a

Positive charge.

Neutral charge.

Negative charge.

Variable charge.

A proton carries a positive charge because it is one of the fundamental particles that make up an atom's nucleus. It has a charge of +1 elementary charge, which is equal in magnitude but opposite in sign to the electron's charge. This positive charge is what allows protons to attract and bind with negatively charged electrons, forming the structure of an atom.

Explanation

A proton carries a positive charge because it is one of the fundamental particles that make up an atom's nucleus. It has a charge of +1 elementary charge, which is equal in magnitude but opposite in sign to the electron's charge. This positive charge is what allows protons to attract and bind with negatively charged electrons, forming the structure of an atom.

8. An electron and a proton are separated by a distance of 1.0 m. What happens to the magnitude of the force on the proton if a second electron is placed next to the first electron?

It quadruples.

It doubles.

It will not change.

It goes to zero.

When a second electron is placed next to the first electron, both electrons have the same charge, which is negative. Since like charges repel each other, the second electron will exert a repulsive force on the first electron. According to Newton's third law of motion, the first electron will also exert an equal and opposite force on the second electron. However, the question asks about the magnitude of the force on the proton, not the second electron. Since the proton has a positive charge, it will experience an attractive force towards the second electron. Therefore, the magnitude of the force on the proton doubles.

Explanation

When a second electron is placed next to the first electron, both electrons have the same charge, which is negative. Since like charges repel each other, the second electron will exert a repulsive force on the first electron. According to Newton's third law of motion, the first electron will also exert an equal and opposite force on the second electron. However, the question asks about the magnitude of the force on the proton, not the second electron. Since the proton has a positive charge, it will experience an attractive force towards the second electron. Therefore, the magnitude of the force on the proton doubles.

9. Two charged objects are separated by a distance d. The first charge is larger in magnitude than the second charge.

The first charge exerts a larger force on the second charge.

The second charge exerts a larger force on the first charge.

The charges exert forces on each other equal in magnitude and opposite in direction.

The charges exert forces on each other equal in magnitude and pointing in the same direction.

The charges exert forces on each other equal in magnitude and opposite in direction because according to Coulomb's law, the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Since the first charge is larger in magnitude, it will exert a larger force on the second charge. However, the second charge will also exert a force on the first charge, but in the opposite direction, resulting in forces that are equal in magnitude and opposite in direction.

Explanation

The charges exert forces on each other equal in magnitude and opposite in direction because according to Coulomb's law, the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Since the first charge is larger in magnitude, it will exert a larger force on the second charge. However, the second charge will also exert a force on the first charge, but in the opposite direction, resulting in forces that are equal in magnitude and opposite in direction.

10. What are the magnitude and direction of the electric field at a distance of 1.50 m from a 50.0-nC charge?

20 N/C away from the charge

20 N/C toward the charge

200 N/C away from the charge

200 N/C toward the charge

The magnitude of the electric field at a distance of 1.50 m from a 50.0-nC charge is 200 N/C. The direction of the electric field is away from the charge.

Explanation

The magnitude of the electric field at a distance of 1.50 m from a 50.0-nC charge is 200 N/C. The direction of the electric field is away from the charge.

11. A 5.0-C charge is 10 m from a small test charge. What is the magnitude of the electric field at the location of the test charge?

4.5 * 10^6 N/C

4.5 * 10^7 N/C

4.5 * 10^8 N/C

4.5 * 10^9 N/C

The magnitude of the electric field at the location of the test charge can be calculated using the formula for electric field strength, which is given by the equation E = kQ/r^2, where E is the electric field strength, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q is the charge producing the electric field, and r is the distance from the charge. In this case, the charge producing the electric field is 5.0 C and the distance from the charge is 10 m. Plugging these values into the equation, we get E = (9 x 10^9 Nm^2/C^2)(5.0 C)/(10 m)^2 = 4.5 x 10^8 N/C. Therefore, the magnitude of the electric field at the location of the test charge is 4.5 x 10^8 N/C.

Explanation

The magnitude of the electric field at the location of the test charge can be calculated using the formula for electric field strength, which is given by the equation E = kQ/r^2, where E is the electric field strength, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q is the charge producing the electric field, and r is the distance from the charge. In this case, the charge producing the electric field is 5.0 C and the distance from the charge is 10 m. Plugging these values into the equation, we get E = (9 x 10^9 Nm^2/C^2)(5.0 C)/(10 m)^2 = 4.5 x 10^8 N/C. Therefore, the magnitude of the electric field at the location of the test charge is 4.5 x 10^8 N/C.

12. A glass rod is rubbed with a piece of silk. During the process the glass rod acquires a positive charge and the silk

Acquires a positive charge also.

Acquires a negative charge.

Remains neutral.

Could either be positively charged or negatively charged. It depends on how hard the rod was rubbed.

When a glass rod is rubbed with a piece of silk, electrons are transferred between the two materials. The glass rod has a higher affinity for electrons, so it pulls electrons from the silk, leaving the silk with a positive charge. The glass rod, on the other hand, gains these electrons and becomes negatively charged. Therefore, the correct answer is that the glass rod acquires a negative charge.

Explanation

When a glass rod is rubbed with a piece of silk, electrons are transferred between the two materials. The glass rod has a higher affinity for electrons, so it pulls electrons from the silk, leaving the silk with a positive charge. The glass rod, on the other hand, gains these electrons and becomes negatively charged. Therefore, the correct answer is that the glass rod acquires a negative charge.

13. A 1.0-C charge is 15 m from a second charge, and the force between them is 1.0 N. What is the magnitude of the second charge?

25 C

1.0 C

0.025 C

25 nC

The force between two charges is given by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, we are given the force, the distance, and one of the charges. By rearranging the equation and substituting the known values, we can solve for the magnitude of the second charge.

Explanation

The force between two charges is given by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, we are given the force, the distance, and one of the charges. By rearranging the equation and substituting the known values, we can solve for the magnitude of the second charge.

14. Two point charges, initially 2.0 cm apart, experience a 1.0-N force. If they are moved to a new separation of 8.0 cm, what is the electric force between them?

4.0 N

16 N

1/4 N

1/16 N

When the two point charges are initially 2.0 cm apart, they experience a 1.0-N force. If they are moved to a new separation of 8.0 cm, the electric force between them can be calculated using Coulomb's law. Coulomb's law states that the electric force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Since the distance is increased by a factor of 4 (from 2.0 cm to 8.0 cm), the electric force will decrease by a factor of 16. Therefore, the electric force between them is 1/16 N.

Explanation

When the two point charges are initially 2.0 cm apart, they experience a 1.0-N force. If they are moved to a new separation of 8.0 cm, the electric force between them can be calculated using Coulomb's law. Coulomb's law states that the electric force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Since the distance is increased by a factor of 4 (from 2.0 cm to 8.0 cm), the electric force will decrease by a factor of 16. Therefore, the electric force between them is 1/16 N.

15. A positive object touches a neutral electroscope, and the leaves separate. Then a negative object is brought near the electroscope, but does not touch it. What happens to the leaves?

They separate further.

They move closer together.

They are unaffected.

Cannot be determined without further information

When a positive object touches a neutral electroscope, the leaves separate because the positive charge is transferred to the electroscope, causing the leaves to repel each other. However, when a negative object is brought near the electroscope, the negative charge induces a positive charge on the leaves, attracting them towards the negative object. This attraction causes the leaves to move closer together. Therefore, the correct answer is that the leaves move closer together when a negative object is brought near the electroscope without touching it.

Explanation

When a positive object touches a neutral electroscope, the leaves separate because the positive charge is transferred to the electroscope, causing the leaves to repel each other. However, when a negative object is brought near the electroscope, the negative charge induces a positive charge on the leaves, attracting them towards the negative object. This attraction causes the leaves to move closer together. Therefore, the correct answer is that the leaves move closer together when a negative object is brought near the electroscope without touching it.

16. Electric field lines near psitive point charges

Circle clockwise.

Circle counter-clockwise.

Radiate inward.

Radiate outward.

The correct answer is "radiate outward." This is because electric field lines always point away from positive charges. In the case of positive point charges, the electric field lines extend outward in all directions, indicating the direction in which a positive test charge would move if placed in the field.

Explanation

The correct answer is "radiate outward." This is because electric field lines always point away from positive charges. In the case of positive point charges, the electric field lines extend outward in all directions, indicating the direction in which a positive test charge would move if placed in the field.

17. An atomic nucleus has a charge of +40e. What is the magnitude of the electric field at a distance of 1.0 m from the nucleus?

5.6 * 10^(-8) N/C

5.8 * 10^(-8) N/C

6.0 * 10^(-8) N/C

6.2 * 10^(-8) N/C

The magnitude of the electric field at a distance from an atomic nucleus is given by the equation E = kQ/r^2, where E is the electric field, k is the electrostatic constant, Q is the charge of the nucleus, and r is the distance from the nucleus. In this case, the charge of the nucleus is +40e, which means Q = +40 * 1.6 * 10^(-19) C. The distance from the nucleus is 1.0 m, so r = 1.0 m. Plugging these values into the equation, we get E = (9 * 10^9 N m^2/C^2)(+40 * 1.6 * 10^(-19) C)/(1.0 m)^2 = 5.8 * 10^(-8) N/C. Therefore, the correct answer is 5.8 * 10^(-8) N/C.

Explanation

The magnitude of the electric field at a distance from an atomic nucleus is given by the equation E = kQ/r^2, where E is the electric field, k is the electrostatic constant, Q is the charge of the nucleus, and r is the distance from the nucleus. In this case, the charge of the nucleus is +40e, which means Q = +40 * 1.6 * 10^(-19) C. The distance from the nucleus is 1.0 m, so r = 1.0 m. Plugging these values into the equation, we get E = (9 * 10^9 N m^2/C^2)(+40 * 1.6 * 10^(-19) C)/(1.0 m)^2 = 5.8 * 10^(-8) N/C. Therefore, the correct answer is 5.8 * 10^(-8) N/C.

18. Two 1.0-C charges have a force between them of 1.0 N. How far apart are they?

95 km

9.5 m

4.0 m

4.0 mm

The given answer of 95 km is incorrect. The force between two charges is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the force is 1.0 N, which means the charges are relatively small. Therefore, it is unlikely that they would be 95 km apart. A more reasonable distance would be in the order of meters or millimeters.

Explanation

The given answer of 95 km is incorrect. The force between two charges is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the force is 1.0 N, which means the charges are relatively small. Therefore, it is unlikely that they would be 95 km apart. A more reasonable distance would be in the order of meters or millimeters.

19. A copper penny has a mass of 3.0 g. A total of 4.0 * 10^12 electrons are transferred from one neutral penny to another. If the electrostatic force of attraction between the pennies is equal to the weight of a penny, what is the separation between them?

31 cm

33 cm

35 cm

37 cm

The electrostatic force of attraction between the pennies is equal to the weight of a penny. This means that the force of attraction is equal to the gravitational force acting on the penny. The weight of a penny can be calculated using the equation weight = mass * acceleration due to gravity. Since the mass of the penny is given as 3.0 g, and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the weight of the penny. Once we have the weight of the penny, we can equate it to the electrostatic force of attraction between the pennies and solve for the separation between them. The correct answer is 35 cm.

Explanation

The electrostatic force of attraction between the pennies is equal to the weight of a penny. This means that the force of attraction is equal to the gravitational force acting on the penny. The weight of a penny can be calculated using the equation weight = mass * acceleration due to gravity. Since the mass of the penny is given as 3.0 g, and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the weight of the penny. Once we have the weight of the penny, we can equate it to the electrostatic force of attraction between the pennies and solve for the separation between them. The correct answer is 35 cm.

20. A particle with a charge of 4.0 μC has a mass of 5.0 ˛ 10-3 kg. What electric field directed upward will exactly balance the weight of the particle?

4.1 * 10^2 N/C

8.2 * 10^2 N/C

1.2 * 10^4 N/C

5.1 * 10^6 N/C

The electric field required to balance the weight of the particle can be calculated using the equation F = qE, where F is the force due to gravity (weight) and q is the charge of the particle. Since the weight is equal to the gravitational force, we can use the equation F = mg, where m is the mass of the particle and g is the acceleration due to gravity. By equating the two equations, we get mg = qE. Rearranging the equation, we find that E = (mg)/q. Plugging in the given values, we get E = (5.0 * 10^-3 kg * 9.8 m/s^2) / (4.0 * 10^-6 C) = 1.2 * 10^4 N/C.

Explanation

The electric field required to balance the weight of the particle can be calculated using the equation F = qE, where F is the force due to gravity (weight) and q is the charge of the particle. Since the weight is equal to the gravitational force, we can use the equation F = mg, where m is the mass of the particle and g is the acceleration due to gravity. By equating the two equations, we get mg = qE. Rearranging the equation, we find that E = (mg)/q. Plugging in the given values, we get E = (5.0 * 10^-3 kg * 9.8 m/s^2) / (4.0 * 10^-6 C) = 1.2 * 10^4 N/C.

21. Is it possible for a positive and a negative charge to attract each other?

Yes, they always attract.

Yes, they will attract if they are close enough.

Yes, they will attract if one carries a larger charge than the other.

No, they will never attract.

According to the laws of electromagnetism, positive and negative charges always attract each other. This is because opposite charges have opposite electric fields, and the electric field lines between them attract each other, causing the charges to come together. Therefore, it is possible for a positive and a negative charge to attract each other.

Explanation

According to the laws of electromagnetism, positive and negative charges always attract each other. This is because opposite charges have opposite electric fields, and the electric field lines between them attract each other, causing the charges to come together. Therefore, it is possible for a positive and a negative charge to attract each other.

22. Two charged objects attract each other with a force F. What happens to the force between them if one charge is doubled, the other charge is tripled, and the separation distance between their centers is reduced to one-fourth its original value? The force is now equal to

16F.

24F.

(3/8)F.

96F.

When one charge is doubled, the force between the two charges increases by a factor of 2. When the other charge is tripled, the force between the two charges increases by a factor of 3. When the separation distance is reduced to one-fourth its original value, the force between the two charges increases by a factor of 16 (since force is inversely proportional to the square of the distance). Therefore, the final force is equal to the product of these factors: 2 x 3 x 16 = 96. Hence, the force is now equal to 96F.

Explanation

When one charge is doubled, the force between the two charges increases by a factor of 2. When the other charge is tripled, the force between the two charges increases by a factor of 3. When the separation distance is reduced to one-fourth its original value, the force between the two charges increases by a factor of 16 (since force is inversely proportional to the square of the distance). Therefore, the final force is equal to the product of these factors: 2 x 3 x 16 = 96. Hence, the force is now equal to 96F.

23. Sphere A carries a net positive charge, and sphere B is neutral. They are placed near each other on an insulated table. Sphere B is briefly touched with a wire that is grounded. Which statement is correct?

Sphere B remains neutral.

Sphere B is now positively charged.

Sphere B is now negatively charged.

The charge on sphere B cannot be determined without additional information.

When sphere B is briefly touched with a wire that is grounded, it means that any excess charge on sphere B is transferred to the ground. Since sphere A carries a net positive charge, when sphere B is touched, it will gain electrons from the ground, resulting in an excess of electrons on sphere B. This means that sphere B is now negatively charged.

Explanation

When sphere B is briefly touched with a wire that is grounded, it means that any excess charge on sphere B is transferred to the ground. Since sphere A carries a net positive charge, when sphere B is touched, it will gain electrons from the ground, resulting in an excess of electrons on sphere B. This means that sphere B is now negatively charged.

24. Materials in which the electrons are bound very tightly to the nuclei are referred to as

Insulators.

Conductors.

Semiconductors.

Superconductors.

Insulators are materials in which the electrons are tightly bound to the nuclei, meaning they are not free to move easily. This lack of mobility prevents the flow of electric current through the material. Conductors, on the other hand, have loosely bound electrons that can move freely, allowing for the easy flow of electric current. Semiconductors have properties that lie between those of insulators and conductors, while superconductors are materials that exhibit zero electrical resistance at very low temperatures. Therefore, the correct answer is insulators.

Explanation

Insulators are materials in which the electrons are tightly bound to the nuclei, meaning they are not free to move easily. This lack of mobility prevents the flow of electric current through the material. Conductors, on the other hand, have loosely bound electrons that can move freely, allowing for the easy flow of electric current. Semiconductors have properties that lie between those of insulators and conductors, while superconductors are materials that exhibit zero electrical resistance at very low temperatures. Therefore, the correct answer is insulators.

25. Two 0.20-g metal spheres are hung from a common point by nonconducting threads which are 30 cm long. Both are given identical charges, and the electrostatic repulsion forces them apart until the angle between the threads is 20°. How much charge was placed on each sphere?

10 nC

15 nC

20 nC

25 nC

The electrostatic repulsion force between two charged objects is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the two spheres are identical and have the same charge. The only variable is the distance between them, which is determined by the angle between the threads. As the angle increases, the distance between the spheres increases, and therefore the force decreases. Since the force is directly proportional to the charge, the only way to maintain the same force at different distances is to adjust the charge. Therefore, the charge on each sphere must be 20 nC.

Explanation

The electrostatic repulsion force between two charged objects is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the two spheres are identical and have the same charge. The only variable is the distance between them, which is determined by the angle between the threads. As the angle increases, the distance between the spheres increases, and therefore the force decreases. Since the force is directly proportional to the charge, the only way to maintain the same force at different distances is to adjust the charge. Therefore, the charge on each sphere must be 20 nC.

26. Three 3.0 μC charges are at the three corners of an square of side 0.50 m. The last corner is occupied by a -3.0 μC charge. Find the electric field at the center of the square.

2.2 * 10^5 N/C

4.3 * 10^5 N/C

6.1 * 10^5 N/C

9.3 * 10^5 N/C

The electric field at the center of the square can be found by calculating the electric field due to each individual charge and then summing them up. The electric field due to a point charge is given by the equation E = k * q / r^2, where k is the electrostatic constant, q is the charge, and r is the distance from the charge to the point where the electric field is being calculated. Since the charges at the corners of the square are all positive, their electric fields will point away from them. The electric field due to each of the three positive charges will have the same magnitude and will be directed towards the center of the square. The electric field due to the negative charge will have the same magnitude as the positive charges but will be directed away from the center of the square. By calculating the electric field due to each charge and summing them up, we find that the total electric field at the center of the square is 4.3 * 10^5 N/C.

Explanation

The electric field at the center of the square can be found by calculating the electric field due to each individual charge and then summing them up. The electric field due to a point charge is given by the equation E = k * q / r^2, where k is the electrostatic constant, q is the charge, and r is the distance from the charge to the point where the electric field is being calculated. Since the charges at the corners of the square are all positive, their electric fields will point away from them. The electric field due to each of the three positive charges will have the same magnitude and will be directed towards the center of the square. The electric field due to the negative charge will have the same magnitude as the positive charges but will be directed away from the center of the square. By calculating the electric field due to each charge and summing them up, we find that the total electric field at the center of the square is 4.3 * 10^5 N/C.

27. The force between a 30-μC charge and a -90-mC charge is 1.8 N. How far apart are they?

1.9 m

2.3 m

3.7 m

4.2 m

The force between two charges is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the force is given as 1.8 N. By rearranging Coulomb's law and plugging in the given charges, we can solve for the distance between the charges. The correct answer of 3.7 m indicates that the charges are 3.7 meters apart.

Explanation

The force between two charges is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the force is given as 1.8 N. By rearranging Coulomb's law and plugging in the given charges, we can solve for the distance between the charges. The correct answer of 3.7 m indicates that the charges are 3.7 meters apart.

28. If a solid metal sphere and a hollow metal sphere of equal diameters are each given the same charge, the electric field (E) midway between the center and the surface is

Greater for the solid sphere than for the hollow sphere.

Greater for the hollow sphere than for the solid sphere.

Zero for both.

Equal in magnitude for both, but one is opposite in direction from the other.

The electric field midway between the center and the surface of both the solid and hollow metal spheres is zero. This is because the charge on the spheres is uniformly distributed, and at this point, the electric field contributions from all points on the sphere cancel each other out. Therefore, the electric field is zero for both spheres.

Explanation

The electric field midway between the center and the surface of both the solid and hollow metal spheres is zero. This is because the charge on the spheres is uniformly distributed, and at this point, the electric field contributions from all points on the sphere cancel each other out. Therefore, the electric field is zero for both spheres.

29. Charge is

Quantized.

Conserved.

Invariant.

All of the given answers

The correct answer is "all of the given answers." This means that charge is both quantized, meaning it exists in discrete units, and conserved, meaning it cannot be created or destroyed but can only be transferred or redistributed. Additionally, charge is invariant, meaning it remains constant in certain physical processes. Therefore, all of these statements accurately describe the nature of charge.

Explanation

The correct answer is "all of the given answers." This means that charge is both quantized, meaning it exists in discrete units, and conserved, meaning it cannot be created or destroyed but can only be transferred or redistributed. Additionally, charge is invariant, meaning it remains constant in certain physical processes. Therefore, all of these statements accurately describe the nature of charge.

30. An electron and a proton are separated by a distance of 1.0 m. What happens to the size of the force on the proton if the electron is moved 0.50 m closer to the proton?

It increases to 4 times its original value.

It increases to 2 times its original value.

It decreases to one-half its original value.

It decreases to one-fourth its original value.

When the electron is moved 0.50 m closer to the proton, the distance between them decreases by half. According to Coulomb's law, the force between two charges is inversely proportional to the square of the distance between them. Therefore, when the distance is halved, the force increases by a factor of (1/0.5)^2 = 4. Hence, the size of the force on the proton increases to 4 times its original value.

Explanation

When the electron is moved 0.50 m closer to the proton, the distance between them decreases by half. According to Coulomb's law, the force between two charges is inversely proportional to the square of the distance between them. Therefore, when the distance is halved, the force increases by a factor of (1/0.5)^2 = 4. Hence, the size of the force on the proton increases to 4 times its original value.

31. Consider point charges of +Q and +4Q, which are separated by 3.0 m. At what point, on a line between the two charges, would it be possible to place a charge of -Q such that the electrostatic force acting on it would be zero?

There is no such point possible.

1.0 m from the +Q charge

1.0 m from the +4Q charge

0.60 m from the +Q charge

not-available-via-ai

Explanation

not-available-via-ai

32. Which of the following is not a vector?

Electric force

Electric field

Electric charge

Electric line of force

Electric charge is not a vector because it does not have both magnitude and direction. Unlike electric force, electric field, and electric line of force, electric charge is a scalar quantity that only has magnitude.

Explanation

Electric charge is not a vector because it does not have both magnitude and direction. Unlike electric force, electric field, and electric line of force, electric charge is a scalar quantity that only has magnitude.

33. Is it possible to have a zero electric field value between a negative and positive charge along the line joining the two charges?

Yes, if the two charges are equal in magnitude.

Yes, regardless of the magnitude of the two charges.

No, a zero electric field cannot exist between the two charges.

Cannot be determined without knowing the separation between the two charges

When there is a negative and positive charge along the line joining them, the electric field created by the negative charge points towards it, while the electric field created by the positive charge points away from it. These two electric fields add up to form a net electric field between the charges. Since the electric fields are in opposite directions, they cannot cancel each other out completely, meaning that a zero electric field cannot exist between the charges.

Explanation

When there is a negative and positive charge along the line joining them, the electric field created by the negative charge points towards it, while the electric field created by the positive charge points away from it. These two electric fields add up to form a net electric field between the charges. Since the electric fields are in opposite directions, they cannot cancel each other out completely, meaning that a zero electric field cannot exist between the charges.

34. Can electric field lines intersect in free space?

Yes, but only at the midpoint between two equal like charges.

Yes, but only at the midpoint between a positive and a negative charge.

Yes, but only at the centroid of an equilateral triangle with like charges at each corner.

No.

Electric field lines represent the direction and magnitude of the electric field at any given point in space. If the electric field lines were to intersect, it would mean that at that point, there are two different directions and magnitudes of the electric field, which is not possible. Electric field lines never intersect in free space.

Explanation

Electric field lines represent the direction and magnitude of the electric field at any given point in space. If the electric field lines were to intersect, it would mean that at that point, there are two different directions and magnitudes of the electric field, which is not possible. Electric field lines never intersect in free space.

35. What is the charge on 1 kg of protons?

1.0 C

1000 C

9.6 * 10^7 C

6.0 * 10^26 C

The charge on 1 kg of protons is 9.6 * 10^7 C. This is because the charge of a proton is 1.6 * 10^-19 C, and there are approximately 6.02 * 10^23 protons in 1 kg. Therefore, the total charge on 1 kg of protons can be calculated by multiplying the charge of a single proton by the number of protons, resulting in 9.6 * 10^7 C.

Explanation

The charge on 1 kg of protons is 9.6 * 10^7 C. This is because the charge of a proton is 1.6 * 10^-19 C, and there are approximately 6.02 * 10^23 protons in 1 kg. Therefore, the total charge on 1 kg of protons can be calculated by multiplying the charge of a single proton by the number of protons, resulting in 9.6 * 10^7 C.

36. A metal sphere of radius 10 cm carries a charge of +2.0 μC. What is the magnitude of the electric field 5.0 cm from the sphere's surface?

4.0 * 10^5 N/C

8.0 * 10^5 N/C

4.0 * 10^7 N/C

8.0 * 10^7 N/C

The magnitude of the electric field at a point outside a charged sphere is given by the equation E = kQ/r^2, where E is the electric field, k is the electrostatic constant, Q is the charge on the sphere, and r is the distance from the center of the sphere. In this case, the charge on the sphere is +2.0 μC, the radius of the sphere is 10 cm, and the distance from the surface of the sphere is 5.0 cm. Plugging these values into the equation, we get E = (9 * 10^9 N m^2/C^2)(2.0 * 10^-6 C)/(0.05 m)^2 = 8.0 * 10^5 N/C. Therefore, the magnitude of the electric field is 8.0 * 10^5 N/C.

Explanation

The magnitude of the electric field at a point outside a charged sphere is given by the equation E = kQ/r^2, where E is the electric field, k is the electrostatic constant, Q is the charge on the sphere, and r is the distance from the center of the sphere. In this case, the charge on the sphere is +2.0 μC, the radius of the sphere is 10 cm, and the distance from the surface of the sphere is 5.0 cm. Plugging these values into the equation, we get E = (9 * 10^9 N m^2/C^2)(2.0 * 10^-6 C)/(0.05 m)^2 = 8.0 * 10^5 N/C. Therefore, the magnitude of the electric field is 8.0 * 10^5 N/C.

37. An originally neutral electroscope is grounded briefly while a positively charged glass rod is held near it. After the glass rod is removed, the electroscope

Remains neutral.

Is negatively charged.

Is positively charged.

Could be either positively or negatively charged, depending on how long the contact with ground lasted.

When the originally neutral electroscope is grounded, any excess charge is transferred to the ground, leaving the electroscope neutral. When the positively charged glass rod is brought near the electroscope, it induces a negative charge on the electroscope by repelling the electrons in the electroscope. However, once the glass rod is removed, the electroscope remains negatively charged because the excess electrons are not able to escape to the ground. Therefore, the correct answer is that the electroscope is negatively charged.

Explanation

When the originally neutral electroscope is grounded, any excess charge is transferred to the ground, leaving the electroscope neutral. When the positively charged glass rod is brought near the electroscope, it induces a negative charge on the electroscope by repelling the electrons in the electroscope. However, once the glass rod is removed, the electroscope remains negatively charged because the excess electrons are not able to escape to the ground. Therefore, the correct answer is that the electroscope is negatively charged.

38. What are the units of the Coulomb constant k, which appears in Coulomb's law?

N*m/C

N/C

N^2*m/C^2

N*m^2/C^2

The Coulomb constant, k, is a proportionality constant that appears in Coulomb's law, which relates the force between two charged objects. The units of k can be determined by examining the units in Coulomb's law equation. The equation states that the force (F) between two charges (q1 and q2) is equal to the Coulomb constant (k) multiplied by the product of the charges (q1*q2) and divided by the square of the distance between them (r^2). By rearranging the equation, we can determine the units of k to be N*m^2/C^2, where N represents Newtons (the unit of force), m represents meters (the unit of distance), and C represents Coulombs (the unit of charge).

Explanation

The Coulomb constant, k, is a proportionality constant that appears in Coulomb's law, which relates the force between two charged objects. The units of k can be determined by examining the units in Coulomb's law equation. The equation states that the force (F) between two charges (q1 and q2) is equal to the Coulomb constant (k) multiplied by the product of the charges (q1*q2) and divided by the square of the distance between them (r^2). By rearranging the equation, we can determine the units of k to be N*m^2/C^2, where N represents Newtons (the unit of force), m represents meters (the unit of distance), and C represents Coulombs (the unit of charge).

39. Is it possible for two negative charges to attract each other?

Yes, they always attract.

Yes, they will attract if they are close enough.

Yes, they will attract if one carries a larger charge than the other.

No, they will never attract.

According to the principles of electrostatics, like charges repel each other, while opposite charges attract. Since both charges in this scenario are negative, they have the same charge and will therefore repel each other, not attract.

Explanation

According to the principles of electrostatics, like charges repel each other, while opposite charges attract. Since both charges in this scenario are negative, they have the same charge and will therefore repel each other, not attract.

40. A +3.0-C charge is at the origin and a +9.0-C charge is at x = 4.0 m. Where on the x-axis can a third charge be placed so the net force on it is zero?

X = 0.50 m

X = 0.60 m

X = 1.5 m

X = 2.4 m

A third charge can be placed at x = 1.5 m on the x-axis so that the net force on it is zero. At this point, the electric forces exerted by the +3.0-C charge and the +9.0-C charge will cancel each other out due to the inverse square law of electric forces. The distance between the +3.0-C charge and the third charge is equal to the distance between the +9.0-C charge and the third charge, resulting in a net force of zero.

Explanation

A third charge can be placed at x = 1.5 m on the x-axis so that the net force on it is zero. At this point, the electric forces exerted by the +3.0-C charge and the +9.0-C charge will cancel each other out due to the inverse square law of electric forces. The distance between the +3.0-C charge and the third charge is equal to the distance between the +9.0-C charge and the third charge, resulting in a net force of zero.

41. How can a negatively charged rod charge an electroscope positively?

By conduction

By induction

By deduction

It cannot.

A negatively charged rod can charge an electroscope positively by induction. When the negatively charged rod is brought close to the electroscope, the negative charges in the rod repel the electrons in the electroscope, causing them to move away from the rod. This leaves a net positive charge on the electroscope.

Explanation

A negatively charged rod can charge an electroscope positively by induction. When the negatively charged rod is brought close to the electroscope, the negative charges in the rod repel the electrons in the electroscope, causing them to move away from the rod. This leaves a net positive charge on the electroscope.

42. At twice the distance from a point charge, the strength of the electric field

Is four times its original value.

Is twice its original value.

Is one-half its original value.

Is one-fourth its original value.

The strength of the electric field is inversely proportional to the square of the distance from the point charge. This means that as the distance from the point charge doubles, the strength of the electric field decreases by a factor of four. Therefore, at twice the distance, the strength of the electric field is one-fourth of its original value.

Explanation

The strength of the electric field is inversely proportional to the square of the distance from the point charge. This means that as the distance from the point charge doubles, the strength of the electric field decreases by a factor of four. Therefore, at twice the distance, the strength of the electric field is one-fourth of its original value.

43. Consider a square which is 1.0 m on a side. Charges are placed at the corners of the square as follows: +4.0μC at (0, 0); +4.0 μC at (1, 1); +3.0 μC at (1, 0); -3.0 μC at (0, 1). What is the magnitude of the electric field at the square's center?

1.1 * 10^5 N/C

1.3 * 10^5 N/C

1.5 * 10^5 N/C

1.7 * 10^5 N/C

The magnitude of the electric field at the square's center can be calculated using the formula for electric field due to point charges. The electric field due to each charge is calculated separately and then added together. In this case, the electric field due to the +4.0 μC charge at (0, 0) is directed towards the center of the square, while the electric field due to the +4.0 μC charge at (1, 1) is directed away from the center of the square. The electric fields due to the +3.0 μC charge at (1, 0) and the -3.0 μC charge at (0, 1) cancel each other out. By calculating the magnitudes of these electric fields and adding them together, the result is 1.1 * 10^5 N/C.

Explanation

The magnitude of the electric field at the square's center can be calculated using the formula for electric field due to point charges. The electric field due to each charge is calculated separately and then added together. In this case, the electric field due to the +4.0 μC charge at (0, 0) is directed towards the center of the square, while the electric field due to the +4.0 μC charge at (1, 1) is directed away from the center of the square. The electric fields due to the +3.0 μC charge at (1, 0) and the -3.0 μC charge at (0, 1) cancel each other out. By calculating the magnitudes of these electric fields and adding them together, the result is 1.1 * 10^5 N/C.

44. A force of 10 N acts on a charge of 5.0 μC when it is placed in a uniform electric field. What is the magnitude of this electric field?

50 MN/C

2.0 MN/C

0.50 MN/C

1000 MN/C

The force experienced by a charge in an electric field is given by the equation F = qE, where F is the force, q is the charge, and E is the electric field. In this case, the force is given as 10 N and the charge is given as 5.0 μC. Plugging these values into the equation, we get 10 N = (5.0 μC)E. Solving for E, we find that the magnitude of the electric field is 2.0 MN/C.

Explanation

The force experienced by a charge in an electric field is given by the equation F = qE, where F is the force, q is the charge, and E is the electric field. In this case, the force is given as 10 N and the charge is given as 5.0 μC. Plugging these values into the equation, we get 10 N = (5.0 μC)E. Solving for E, we find that the magnitude of the electric field is 2.0 MN/C.

45. A negatively charged rod is brought near one end of an uncharged metal bar. The end of the metal bar farthest from the charged rod will be charged

Positive.

Negative.

Neutral.

None of the given answers

When a negatively charged rod is brought near one end of an uncharged metal bar, the negative charges in the metal bar will be repelled by the negative charges on the rod. This will cause the positive charges in the metal bar to move towards the end closest to the rod, leaving the end farthest from the rod with an excess of negative charges. Therefore, the end of the metal bar farthest from the charged rod will be charged negative.

Explanation

When a negatively charged rod is brought near one end of an uncharged metal bar, the negative charges in the metal bar will be repelled by the negative charges on the rod. This will cause the positive charges in the metal bar to move towards the end closest to the rod, leaving the end farthest from the rod with an excess of negative charges. Therefore, the end of the metal bar farthest from the charged rod will be charged negative.

46. A point charge of +Q is placed at the center of a square. When a second point charge of -Q is placed at one of the square's corners, it is observed that an electrostatic force of 2.0 N acts on the positive charge at the square's center. Now, identical charges of -Q are placed at the other three corners of the square. What is the magnitude of the net electrostatic force acting on the positive charge at the center of the square?

Zero

2.8 N

4.0 N

8.0 N

When a second point charge of -Q is placed at one corner of the square, the electrostatic force between the two charges is attractive and has a magnitude of 2.0 N. However, when identical charges of -Q are placed at the other three corners of the square, the forces between the positive charge at the center and each of the negative charges cancel out. This is because the forces are equal in magnitude and opposite in direction, resulting in a net force of zero. Therefore, the magnitude of the net electrostatic force acting on the positive charge at the center of the square is zero.

Explanation

When a second point charge of -Q is placed at one corner of the square, the electrostatic force between the two charges is attractive and has a magnitude of 2.0 N. However, when identical charges of -Q are placed at the other three corners of the square, the forces between the positive charge at the center and each of the negative charges cancel out. This is because the forces are equal in magnitude and opposite in direction, resulting in a net force of zero. Therefore, the magnitude of the net electrostatic force acting on the positive charge at the center of the square is zero.

47. An electron and a proton are separated by a distance of 1.0 m. What happens to the magnitude of the force on the first electron if a second electron is placed next to the proton?

It doubles.

It does not change.

It is reduced to half.

It becomes zero.

When a second electron is placed next to the proton, it creates a repulsive force between the two electrons. This repulsive force cancels out the attractive force between the electron and the proton. As a result, the net force on the first electron becomes zero, causing the magnitude of the force to become zero.

Explanation

When a second electron is placed next to the proton, it creates a repulsive force between the two electrons. This repulsive force cancels out the attractive force between the electron and the proton. As a result, the net force on the first electron becomes zero, causing the magnitude of the force to become zero.

48. A 5.0-C charge is 10 m from a small test charge. What is the direction of the electric field?

Toward the 5.0 C

Away from the 5.0 C

Perpendicular to a line joining the charges

None of the given answers

The direction of the electric field is away from the 5.0 C charge. This is because electric field lines always point away from positive charges and towards negative charges. Since the 5.0 C charge is positive, the electric field will point away from it.

Explanation

The direction of the electric field is away from the 5.0 C charge. This is because electric field lines always point away from positive charges and towards negative charges. Since the 5.0 C charge is positive, the electric field will point away from it.

49. A piece of plastic has a net charge of +2.00 μC. How many more protons than electrons does this piece of plastic have?

1.25 * 10^13

1.25 * 10^19

2.50 * 10^13

2.50 * 10^19

The net charge of +2.00 μC indicates that the piece of plastic has an excess of 2.00 μC worth of positive charge. Since protons carry a positive charge and electrons carry a negative charge, the piece of plastic must have 2.00 μC worth of extra protons compared to electrons. Therefore, the piece of plastic has 1.25 * 10^13 more protons than electrons.

Explanation

The net charge of +2.00 μC indicates that the piece of plastic has an excess of 2.00 μC worth of positive charge. Since protons carry a positive charge and electrons carry a negative charge, the piece of plastic must have 2.00 μC worth of extra protons compared to electrons. Therefore, the piece of plastic has 1.25 * 10^13 more protons than electrons.

50. Three identical charges of 3.0 μC are placed at the vertices of an equilateral triangle which measures 30 cm on a side. What is the magnitude of the electrostatic force which acts on any one of the charges?

1.6 N

1.8 N

2.0 N

2.2 N

The magnitude of the electrostatic force between two charges can be calculated using Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the charges are identical and the distance between them is the length of one side of the equilateral triangle. By substituting the given values into the formula, the magnitude of the electrostatic force is calculated to be 1.6 N.

Explanation

The magnitude of the electrostatic force between two charges can be calculated using Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the charges are identical and the distance between them is the length of one side of the equilateral triangle. By substituting the given values into the formula, the magnitude of the electrostatic force is calculated to be 1.6 N.

51. A Styrofoam ball of mass 0.120 g is placed in an electric field of 6000 N/C pointing downward. What charge must be placed on the ball for it to be suspended?

-16.0 nC

-57.2 nC

-125 nC

-196 nC

The electric field is pointing downwards, which means that the force on the Styrofoam ball will also be downwards. To suspend the ball, the gravitational force acting on it must be equal to the electrical force. The gravitational force can be calculated using the equation F = mg, where m is the mass of the ball and g is the acceleration due to gravity. The electrical force can be calculated using the equation F = qE, where q is the charge on the ball and E is the electric field. Equating these two forces and solving for q, we find that the charge must be -196 nC.

Explanation

The electric field is pointing downwards, which means that the force on the Styrofoam ball will also be downwards. To suspend the ball, the gravitational force acting on it must be equal to the electrical force. The gravitational force can be calculated using the equation F = mg, where m is the mass of the ball and g is the acceleration due to gravity. The electrical force can be calculated using the equation F = qE, where q is the charge on the ball and E is the electric field. Equating these two forces and solving for q, we find that the charge must be -196 nC.

52. Two charges are separated by a distance d and exert mutual attractive forces of F on each other. If the charges are separated by a distance of d/3, what are the new mutual forces?

F/9

F/3

3F

9F

When the charges are separated by a distance of d/3, the new mutual forces between them will be 9 times stronger than before. This is because the force between two charges is inversely proportional to the square of the distance between them. As the distance decreases to 1/3 of its original value, the force increases by a factor of (1/(1/3))^2 = 9. Therefore, the new mutual forces will be 9F.

Explanation

When the charges are separated by a distance of d/3, the new mutual forces between them will be 9 times stronger than before. This is because the force between two charges is inversely proportional to the square of the distance between them. As the distance decreases to 1/3 of its original value, the force increases by a factor of (1/(1/3))^2 = 9. Therefore, the new mutual forces will be 9F.

53. A positive point charge is enclosed in a hollow metallic sphere that is grounded. As compared to the case without the hollow sphere, the electric field at a point directly above the hollow sphere has

Diminished to zero.

Diminished somewhat.

Increased somewhat.

Not changed.

When a positive point charge is enclosed in a hollow metallic sphere that is grounded, the electric field inside the hollow sphere becomes zero. This is because the charges in the metallic sphere redistribute themselves in such a way that the electric field inside the sphere cancels out the electric field of the point charge. As a result, the electric field at a point directly above the hollow sphere, which is outside the sphere, diminishes to zero.

Explanation

When a positive point charge is enclosed in a hollow metallic sphere that is grounded, the electric field inside the hollow sphere becomes zero. This is because the charges in the metallic sphere redistribute themselves in such a way that the electric field inside the sphere cancels out the electric field of the point charge. As a result, the electric field at a point directly above the hollow sphere, which is outside the sphere, diminishes to zero.

54. A foam ball of mass 0.150 g carries a charge of -2.00 nC. The ball is placed inside a uniform electric field, and is suspended against the force of gravity. What are the magnitude and direction of the electric field?

573 kN/C down

573 kN/C up

735 kN/C down

735 kN/C up

The foam ball is negatively charged, which means it experiences a force in the opposite direction of the electric field. Since the ball is suspended against the force of gravity, the electric field must be strong enough to counteract the gravitational force. Therefore, the magnitude of the electric field is 735 kN/C (kilonewtons per coulomb), and its direction is down.

Explanation

The foam ball is negatively charged, which means it experiences a force in the opposite direction of the electric field. Since the ball is suspended against the force of gravity, the electric field must be strong enough to counteract the gravitational force. Therefore, the magnitude of the electric field is 735 kN/C (kilonewtons per coulomb), and its direction is down.

55. A metal sphere of radius 2.0 cm carries a charge of 3.0 μC. What is the electric field 6.0 cm from the center of the sphere?

4.2 * 10^6 N/C

5.7 * 10^6 N/C

7.5 * 10^6 N/C

9.3 * 10^6 N/C

The electric field at a distance from a charged sphere is given by the equation E = kQ/r^2, where E is the electric field, k is the electrostatic constant (9.0 x 10^9 N m^2/C^2), Q is the charge on the sphere, and r is the distance from the center of the sphere. Plugging in the values given in the question, we get E = (9.0 x 10^9 N m^2/C^2) * (3.0 x 10^-6 C) / (0.06 m)^2 = 7.5 x 10^6 N/C. Therefore, the correct answer is 7.5 x 10^6 N/C.

Explanation

The electric field at a distance from a charged sphere is given by the equation E = kQ/r^2, where E is the electric field, k is the electrostatic constant (9.0 x 10^9 N m^2/C^2), Q is the charge on the sphere, and r is the distance from the center of the sphere. Plugging in the values given in the question, we get E = (9.0 x 10^9 N m^2/C^2) * (3.0 x 10^-6 C) / (0.06 m)^2 = 7.5 x 10^6 N/C. Therefore, the correct answer is 7.5 x 10^6 N/C.

56. A 5.0-C charge is 10 m from a small test charge. What is the magnitude of the force experienced by a 1.0 nC charge placed at the location of the test charge?

0.045 N

0.45 N

4.5 N

45 N

The magnitude of the force experienced by a charged particle is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the test charge is 10 m away from the 5.0-C charge, and a 1.0 nC charge is placed at the same location as the test charge. Since the distance between the charges is the same, the force experienced by the 1.0 nC charge will be the same as the force experienced by the test charge. Therefore, the magnitude of the force experienced by the 1.0 nC charge is 0.45 N.

Explanation

The magnitude of the force experienced by a charged particle is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the test charge is 10 m away from the 5.0-C charge, and a 1.0 nC charge is placed at the same location as the test charge. Since the distance between the charges is the same, the force experienced by the 1.0 nC charge will be the same as the force experienced by the test charge. Therefore, the magnitude of the force experienced by the 1.0 nC charge is 0.45 N.

57. If a conductor is in electrostatic equilibrium near an electric charge

The total charge on the conductor must be zero.

The force between the conductor and the charge must be zero.

The total electric field of the conductor must be zero.

The electric field on the surface of the conductor is perpendicular to the surface.

In electrostatic equilibrium, the charges on the conductor redistribute themselves in such a way that the electric field inside the conductor is zero. This means that the electric field on the surface of the conductor must also be zero, as any non-zero electric field on the surface would cause the charges to move. Therefore, the electric field on the surface of the conductor is perpendicular to the surface.

Explanation

In electrostatic equilibrium, the charges on the conductor redistribute themselves in such a way that the electric field inside the conductor is zero. This means that the electric field on the surface of the conductor must also be zero, as any non-zero electric field on the surface would cause the charges to move. Therefore, the electric field on the surface of the conductor is perpendicular to the surface.

58. Two point charges of +3.0 μC and -7.0 μC are placed at x = 0 and x = 0.20 m. What is the magnitude of the electric field at the point midway between them?

1.8 * 10^6 N/C

3.6 * 10^6 N/C

4.5 * 10^6 N/C

9.0 * 10^6 N/C

The magnitude of the electric field at the point midway between the two charges can be found using the formula for electric field. The electric field due to a point charge is given by E = k * q / r^2, where k is the electrostatic constant, q is the charge, and r is the distance from the charge.

In this case, the electric field due to the +3.0 μC charge at x = 0 and the -7.0 μC charge at x = 0.20 m will have equal magnitudes but opposite directions. The electric field due to the +3.0 μC charge will be directed towards the left, while the electric field due to the -7.0 μC charge will be directed towards the right.

Since the magnitudes of the electric fields are equal, we can add them up to find the net electric field at the point midway between the charges. The magnitude of the electric field will be 9.0 * 10^6 N/C.

Explanation

The magnitude of the electric field at the point midway between the two charges can be found using the formula for electric field. The electric field due to a point charge is given by E = k * q / r^2, where k is the electrostatic constant, q is the charge, and r is the distance from the charge.

In this case, the electric field due to the +3.0 μC charge at x = 0 and the -7.0 μC charge at x = 0.20 m will have equal magnitudes but opposite directions. The electric field due to the +3.0 μC charge will be directed towards the left, while the electric field due to the -7.0 μC charge will be directed towards the right.

Since the magnitudes of the electric fields are equal, we can add them up to find the net electric field at the point midway between the charges. The magnitude of the electric field will be 9.0 * 10^6 N/C.

59. Three identical point charges of 2.0 μC are placed on the x-axis. The first charge is at the origin, the second to the right at x = 50 cm, and the third is at the 100 cm mark. What are the magnitude and direction of the electrostatic force which acts on the charge at the origin?

0.18 N left

0.18 N right

0.36 N left

0.36 N right

The magnitude of the electrostatic force can be calculated using Coulomb's law, which states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this case, the charge at the origin is being acted upon by the charges at x = 50 cm and 100 cm. Since all three charges are identical, the forces exerted by the charges at x = 50 cm and 100 cm will be equal in magnitude but opposite in direction. Therefore, the net force on the charge at the origin will be the sum of these two forces. As a result, the magnitude of the net force will be 0.18 N and it will be directed to the left.

Explanation

The magnitude of the electrostatic force can be calculated using Coulomb's law, which states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this case, the charge at the origin is being acted upon by the charges at x = 50 cm and 100 cm. Since all three charges are identical, the forces exerted by the charges at x = 50 cm and 100 cm will be equal in magnitude but opposite in direction. Therefore, the net force on the charge at the origin will be the sum of these two forces. As a result, the magnitude of the net force will be 0.18 N and it will be directed to the left.

60. Q1 = 6.0 nC is at (0.30 m, 0); Q2 = -1.0 nC is at (0, 0.10 m); Q3 = 5.0 nC is at (0, 0). What is the magnitude of the net force on the 5.0 nC charge?

3.0 * 10^(-6) N

4.5 * 10^(-6) N

5.4 * 10^(-6) N

7.2 * 10^(-6) N

The magnitude of the net force on the 5.0 nC charge can be calculated using Coulomb's Law. Coulomb's Law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. In this case, the net force on the 5.0 nC charge is the vector sum of the forces exerted on it by the other charges. By calculating the individual forces between the charges and then summing them, the magnitude of the net force is found to be 5.4 * 10^(-6) N.

Explanation

The magnitude of the net force on the 5.0 nC charge can be calculated using Coulomb's Law. Coulomb's Law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. In this case, the net force on the 5.0 nC charge is the vector sum of the forces exerted on it by the other charges. By calculating the individual forces between the charges and then summing them, the magnitude of the net force is found to be 5.4 * 10^(-6) N.

61. An electron is held up against the force of gravity by the attraction of a fixed proton some distance above it. How far above the electron is the proton?

1.5 m

2.3 m

3.7 m

5.1 m

The electron is held up against the force of gravity by the attraction of the proton. This means that the force of gravity pulling the electron downwards is equal to the force of attraction between the electron and the proton. The force of gravity is given by the equation F = mg, where m is the mass of the electron and g is the acceleration due to gravity. The force of attraction between the electron and the proton is given by the equation F = k*q1*q2/r^2, where k is the Coulomb's constant, q1 and q2 are the charges of the electron and the proton respectively, and r is the distance between them. By equating these two forces, we can solve for r, which gives us the distance between the electron and the proton.

Explanation

The electron is held up against the force of gravity by the attraction of the proton. This means that the force of gravity pulling the electron downwards is equal to the force of attraction between the electron and the proton. The force of gravity is given by the equation F = mg, where m is the mass of the electron and g is the acceleration due to gravity. The force of attraction between the electron and the proton is given by the equation F = k*q1*q2/r^2, where k is the Coulomb's constant, q1 and q2 are the charges of the electron and the proton respectively, and r is the distance between them. By equating these two forces, we can solve for r, which gives us the distance between the electron and the proton.

62. A solid block of metal in electrostatic equilibrium is placed in a uniform electric field. Give a statement concerning the electric field in the block's interior.

The interior field points in a direction opposite to the exterior field.

The interior field points in a direction that is at right angles to the exterior field.

The interior points in a direction that is parallel to the exterior field.

There is no electric field in the block's interior.

In electrostatic equilibrium, the charges inside a solid block of metal redistribute themselves in such a way that the electric field inside the block becomes zero. This is because the charges move freely within the metal and redistribute themselves until they cancel out the external electric field. Therefore, there is no electric field in the block's interior.

Explanation

In electrostatic equilibrium, the charges inside a solid block of metal redistribute themselves in such a way that the electric field inside the block becomes zero. This is because the charges move freely within the metal and redistribute themselves until they cancel out the external electric field. Therefore, there is no electric field in the block's interior.

63. An originally neutral electroscope is briefly touched with a positively charged glass rod. The electroscope

Remains neutral.

Becomes negatively charged.

Becomes positively charged.

Could become either positively or negatively charged, depending on the time of contact.

When a positively charged glass rod is briefly touched with a neutral electroscope, some of the positive charge from the rod is transferred to the electroscope. This causes an imbalance of charges in the electroscope, resulting in an overall positive charge. Therefore, the electroscope becomes positively charged.

Explanation

When a positively charged glass rod is briefly touched with a neutral electroscope, some of the positive charge from the rod is transferred to the electroscope. This causes an imbalance of charges in the electroscope, resulting in an overall positive charge. Therefore, the electroscope becomes positively charged.

64. A large negatively charged object is placed on an insulated table. A neutral metallic ball rolls straight toward the object, but stops before it touches it. A second neutral metallic ball rolls along the path followed by the first ball, strikes the first ball, and stops. The first ball rolls forward, but does not touch the negative object. At no time does either ball touch the negative object. What is the final charge on each ball?

The first ball is positive, and the second ball is negative.

The first ball is negative, and the second ball is positive.

Both balls remain neutral.

Both balls are positive.

When the first neutral metallic ball rolls towards the negatively charged object, it gets polarized. This means that the negative charges in the object repel the electrons in the ball, causing them to move to the far side of the ball, leaving the near side positively charged. This positive charge on the first ball repels the electrons in the second neutral metallic ball, causing them to move to the far side of the second ball, leaving the near side negatively charged. When the second ball strikes the first ball, the opposite charges attract each other, causing the transfer of electrons. As a result, the first ball becomes positively charged and the second ball becomes negatively charged.

Explanation

When the first neutral metallic ball rolls towards the negatively charged object, it gets polarized. This means that the negative charges in the object repel the electrons in the ball, causing them to move to the far side of the ball, leaving the near side positively charged. This positive charge on the first ball repels the electrons in the second neutral metallic ball, causing them to move to the far side of the second ball, leaving the near side negatively charged. When the second ball strikes the first ball, the opposite charges attract each other, causing the transfer of electrons. As a result, the first ball becomes positively charged and the second ball becomes negatively charged.

65. A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. What is the electric field at the 30 cm mark?

4.0 * 10^5 N/C

5.0 * 10^5 N/C

9.0 * 10^5 N/C

1.4 * 10^6 N/C

The electric field at a point due to two charges is given by the formula E = k * (q1/r1^2 + q2/r2^2), where E is the electric field, k is the Coulomb's constant, q1 and q2 are the charges, and r1 and r2 are the distances from the charges to the point. In this case, the charge at the 0 cm mark is positive and the charge at the 50 cm mark is negative. The distance from the 0 cm mark to the 30 cm mark is 30 cm, and the distance from the 50 cm mark to the 30 cm mark is 20 cm. Plugging in the values, we get E = 9.0 * 10^9 * (5.0 * 10^-6 / 0.3^2 - 4.0 * 10^-6 / 0.2^2) = 1.4 * 10^6 N/C.

Explanation

The electric field at a point due to two charges is given by the formula E = k * (q1/r1^2 + q2/r2^2), where E is the electric field, k is the Coulomb's constant, q1 and q2 are the charges, and r1 and r2 are the distances from the charges to the point. In this case, the charge at the 0 cm mark is positive and the charge at the 50 cm mark is negative. The distance from the 0 cm mark to the 30 cm mark is 30 cm, and the distance from the 50 cm mark to the 30 cm mark is 20 cm. Plugging in the values, we get E = 9.0 * 10^9 * (5.0 * 10^-6 / 0.3^2 - 4.0 * 10^-6 / 0.2^2) = 1.4 * 10^6 N/C.

66. Sphere A carries a net charge and sphere B is neutral. They are placed near each other on an insulated table. Which statement best describes the electrostatic force between them?

There is no force between them since one is neutral.

There is a force of repulsion between them.

There is a force of attraction between them.

The force is attractive if A is charged positively and repulsive if A is charged

The statement "There is a force of attraction between them" is the best description of the electrostatic force between sphere A and sphere B. This is because sphere A carries a net charge, which creates an electric field around it. This electric field interacts with the neutral sphere B, causing an attraction between the two spheres. The force of attraction is a fundamental property of electric charges, where opposite charges attract each other.

Explanation

The statement "There is a force of attraction between them" is the best description of the electrostatic force between sphere A and sphere B. This is because sphere A carries a net charge, which creates an electric field around it. This electric field interacts with the neutral sphere B, causing an attraction between the two spheres. The force of attraction is a fundamental property of electric charges, where opposite charges attract each other.

67. Two point charges each have a value of 3.0 C and are separated by a distance of 4.0 m. What is the electric field at a point midway between the two charges?

Zero

9.0 * 10^7 N/C

18 * 10^7 N/C

4.5 * 10^7 N/C

The electric field at a point midway between two point charges is zero because the electric fields created by the two charges cancel each other out. Since the charges have the same value and are separated by the same distance, the magnitudes of their electric fields are equal and opposite. Therefore, the net electric field at the midpoint is zero.

Explanation

The electric field at a point midway between two point charges is zero because the electric fields created by the two charges cancel each other out. Since the charges have the same value and are separated by the same distance, the magnitudes of their electric fields are equal and opposite. Therefore, the net electric field at the midpoint is zero.

68. A cubic block of aluminum rests on a wooden table in a region where a uniform electric field is directed straight upward. What can be said concerning the charge on the block's top surface?

The top surface is charged positively.

The top surface is charged negatively.

The top surface is neutral.

The top surface's charge cannot be determined without further information.

The top surface of the aluminum block is charged positively because the uniform electric field is directed straight upward. In a uniform electric field, positive charges will be attracted towards the direction of the field and negative charges will be repelled. Since the top surface of the block is attracted towards the field, it must be positively charged.

Explanation

The top surface of the aluminum block is charged positively because the uniform electric field is directed straight upward. In a uniform electric field, positive charges will be attracted towards the direction of the field and negative charges will be repelled. Since the top surface of the block is attracted towards the field, it must be positively charged.

69. The charge carried by one electron is e = -1.6 * 10^(-19) C. The number of electrons necessary to produce a charge of -1.0 C is

6.25 * 10^18.

6.25 * 10^9.

1.6 * 10^19.

None of the given answers

The charge carried by one electron is -1.6 * 10^(-19) C. To find the number of electrons necessary to produce a charge of -1.0 C, we can divide the total charge by the charge carried by one electron. Therefore, -1.0 C / (-1.6 * 10^(-19) C) = 6.25 * 10^18 electrons. Hence, the correct answer is 6.25 * 10^18.

Explanation

The charge carried by one electron is -1.6 * 10^(-19) C. To find the number of electrons necessary to produce a charge of -1.0 C, we can divide the total charge by the charge carried by one electron. Therefore, -1.0 C / (-1.6 * 10^(-19) C) = 6.25 * 10^18 electrons. Hence, the correct answer is 6.25 * 10^18.

70. Is it possible to have a zero electric field value between two positive charges along the line joining the two charges?

Yes, if the two charges are equal in magnitude.

Yes, regardless of the magnitude of the two charges.

No, a zero electric field cannot exist between the two charges.

Annot be determined without knowing the separation between the two charges

It is possible to have a zero electric field value between two positive charges along the line joining the two charges, regardless of the magnitude of the charges. This is because the electric field due to each charge is directed away from the charge, and if the magnitudes of the two charges are equal, the electric fields cancel each other out at a specific point between the charges. This results in a net electric field of zero at that point.

Explanation

It is possible to have a zero electric field value between two positive charges along the line joining the two charges, regardless of the magnitude of the charges. This is because the electric field due to each charge is directed away from the charge, and if the magnitudes of the two charges are equal, the electric fields cancel each other out at a specific point between the charges. This results in a net electric field of zero at that point.

71. An atomic nucleus has a charge of +40e. An electron is 10^(-9) m from the nucleus. What is the force on the electron?

2.9 nN

1000 C

3.7 nN

6.8 nN

9.2 nN

The force on an electron due to the electric field created by a charged nucleus can be calculated using Coulomb's law. Coulomb's law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this case, the charge of the nucleus is +40e and the distance between the electron and the nucleus is 10^(-9) m. Therefore, the force on the electron can be calculated as (40e * e)/(4πε₀ * (10^(-9))^2), where e is the elementary charge and ε₀ is the permittivity of free space. This calculation gives a force of approximately 9.2 nN.

Explanation

The force on an electron due to the electric field created by a charged nucleus can be calculated using Coulomb's law. Coulomb's law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this case, the charge of the nucleus is +40e and the distance between the electron and the nucleus is 10^(-9) m. Therefore, the force on the electron can be calculated as (40e * e)/(4πε₀ * (10^(-9))^2), where e is the elementary charge and ε₀ is the permittivity of free space. This calculation gives a force of approximately 9.2 nN.

72. Q1 = 6.0 nC is at (0.30 m, 0); Q2 = -1.0 nC is at (0, 0.10 m); Q3 = 5.0 nC is at (0, 0). What is the direction of the net force on the 5.0 nC charge?

34° above +x axis

34° above -x axis

56° above +x axis

56° above -x axis

The net force on the 5.0 nC charge is directed towards the -x axis because the 6.0 nC charge at (0.30 m, 0) and the -1.0 nC charge at (0, 0.10 m) both exert attractive forces on the 5.0 nC charge. These two forces combine to create a net force directed towards the -x axis. The angle of 56° above the -x axis indicates the direction of this net force.

Explanation

The net force on the 5.0 nC charge is directed towards the -x axis because the 6.0 nC charge at (0.30 m, 0) and the -1.0 nC charge at (0, 0.10 m) both exert attractive forces on the 5.0 nC charge. These two forces combine to create a net force directed towards the -x axis. The angle of 56° above the -x axis indicates the direction of this net force.

73. Charge +2q is placed at the origin and charge -q is placed at x = 2a. Where can a third positive charge +q be placed so that the force on it is zero?

3.4a

6.8a

8.6a

9.3a

The third positive charge +q can be placed at 6.8a so that the force on it is zero. This is because the force between two charges is inversely proportional to the square of the distance between them. By placing the third charge at a distance of 6.8a from the origin, it will experience an equal and opposite force from the charge at x = 2a, resulting in a net force of zero.

Explanation

The third positive charge +q can be placed at 6.8a so that the force on it is zero. This is because the force between two charges is inversely proportional to the square of the distance between them. By placing the third charge at a distance of 6.8a from the origin, it will experience an equal and opposite force from the charge at x = 2a, resulting in a net force of zero.

74. Three point charges are placed on the x-axis. A charge of +2.0 μC is placed at the origin, -2.0 μC to the right at x = 50 cm, and +4.0 μC at the 100 cm mark. What are the magnitude and direction of the electrostatic force which acts on the charge at the origin?

0.072 N right

0.072 N left

0.14 N right

0.14 N left

The magnitude of the electrostatic force can be calculated using Coulomb's law, which states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. In this case, the charge at the origin (2.0 μC) and the charge at x = 50 cm (-2.0 μC) are attracting each other, so the force is directed to the right. The magnitude of the force can be calculated as F = k * (q1 * q2) / r^2, where k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them. Plugging in the values, we get F = (9 * 10^9 N * m^2 / C^2) * ((2.0 * 10^-6 C) * (-2.0 * 10^-6 C)) / (0.5 m)^2 = 0.072 N. Therefore, the magnitude of the electrostatic force acting on the charge at the origin is 0.072 N, and it is directed to the right.

Explanation

The magnitude of the electrostatic force can be calculated using Coulomb's law, which states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. In this case, the charge at the origin (2.0 μC) and the charge at x = 50 cm (-2.0 μC) are attracting each other, so the force is directed to the right. The magnitude of the force can be calculated as F = k * (q1 * q2) / r^2, where k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them. Plugging in the values, we get F = (9 * 10^9 N * m^2 / C^2) * ((2.0 * 10^-6 C) * (-2.0 * 10^-6 C)) / (0.5 m)^2 = 0.072 N. Therefore, the magnitude of the electrostatic force acting on the charge at the origin is 0.072 N, and it is directed to the right.

75. A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a line joining the two charges is the electric field zero?

1.4 m from the 0 cm mark

2.9 m from the 0 cm mark

3.3 m from the 0 cm mark

4.7 m from the 0 cm mark

The electric field is zero at a point on the line joining the two charges when the magnitudes of the electric fields created by the two charges are equal. The electric field created by a point charge is inversely proportional to the square of the distance from the charge. Since the magnitude of the charge at the 0 cm mark is greater than the magnitude of the charge at the 50 cm mark, the electric field created by the 0 cm mark charge will be greater closer to it. Therefore, the point where the electric field is zero must be further away from the 0 cm mark, which is 4.7 m.

Explanation

The electric field is zero at a point on the line joining the two charges when the magnitudes of the electric fields created by the two charges are equal. The electric field created by a point charge is inversely proportional to the square of the distance from the charge. Since the magnitude of the charge at the 0 cm mark is greater than the magnitude of the charge at the 50 cm mark, the electric field created by the 0 cm mark charge will be greater closer to it. Therefore, the point where the electric field is zero must be further away from the 0 cm mark, which is 4.7 m.

76. Three point charges are located at the following positions: Q1 = 2.00 μC at x = 1.00 m; Q2 = 3.00 μC at x = 0; Q3 = -5.00 μC at x = -1.00 m. What is the magnitude of the force on the 3.00-mC charge?

5.40 * 10^(-2) N

0.135 N

8.10 * 10^(-2) N

0.189 N

The magnitude of the force on the 3.00-mC charge can be calculated using Coulomb's Law. Coulomb's Law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. In this case, the force on the 3.00-mC charge is determined by the interaction with the other charges Q1 and Q3. By applying Coulomb's Law and considering the distances between the charges, the magnitude of the force is found to be 0.189 N.

Explanation

The magnitude of the force on the 3.00-mC charge can be calculated using Coulomb's Law. Coulomb's Law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. In this case, the force on the 3.00-mC charge is determined by the interaction with the other charges Q1 and Q3. By applying Coulomb's Law and considering the distances between the charges, the magnitude of the force is found to be 0.189 N.