Chapter 17: Electric Potential

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1/74 Questions

Electric dipoles always consist of two charges that are
- Equal in magnitude; opposite in sign.
- Equal in magnitude; both are negative.
- Equal in magnitude; both are positive.
- Unequal in magnitude; opposite in sign.

About This Quiz

Explore key concepts of electric potential in this focused quiz. It covers vector properties, unit measurements, charge interactions, energy dynamics in electric fields, and charge distribution on conductors. Ideal for students enhancing their understanding in electromagnetism.

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2.

An equipotential surface must be
- Parallel to the electric field at any point.
- Perpendicular to the electric field at any point.
Correct Answer

A. Perpendicular to the electric field at any point.

Explanation

An equipotential surface is a surface in which every point has the same electric potential. Since electric potential is a scalar quantity, it does not have a specific direction. However, electric field is a vector quantity and has a specific direction. In order for every point on an equipotential surface to have the same electric potential, the surface must be perpendicular to the electric field at any point. This is because the electric field lines are always perpendicular to the equipotential surfaces. Therefore, the correct answer is perpendicular to the electric field at any point.

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3.

Which of the following is not a vector?
- Electric force
- Electric field
- Electric potential
- Electric line of force
Correct Answer

A. Electric potential

Explanation

Electric potential is not a vector because it is a scalar quantity. Unlike vectors, which have both magnitude and direction, scalar quantities only have magnitude. Electric potential represents the amount of electric potential energy per unit charge at a given point in an electric field. It does not have a specific direction associated with it, making it a scalar quantity.

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4.

A surface on which all points are at the same potential is referred to as
- A constant electric force surface.
- A constant electric field surface.
- An equipotential surface.
- An equivoltage surface.
Correct Answer

A. An equipotential surface.

Explanation

An equipotential surface is a surface on which all points have the same electric potential. This means that if a charge is placed anywhere on the surface, it will experience no net force. The electric field lines are always perpendicular to an equipotential surface, indicating that the electric field strength is constant at every point on the surface. Therefore, the correct answer is an equipotential surface.

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5.

A dielectric material such as paper is placed between the plates of a capacitor. What happens to the capacitance?
- No change
- Becomes larger
- Becomes smaller
- Becomes infinite
Correct Answer

A. Becomes larger

Explanation

When a dielectric material such as paper is placed between the plates of a capacitor, the capacitance increases. This is because the dielectric material reduces the electric field between the plates, which allows for more charge to be stored on the plates for a given potential difference. As a result, the capacitance of the capacitor increases.

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6.

One electron-volt corresponds to
- 8.0 * 10^(-20) J.
- 1.6 * 10^(-19) J.
- 9.5 * 10^(-17) J.
- 1.9 * 10^(-16) J.
Correct Answer

A. 1.6 * 10^(-19) J.

Explanation

One electron-volt (eV) is a unit of energy equal to the amount of energy gained or lost by a single electron when it moves across an electric potential difference of one volt. The conversion factor between electron-volts and joules (J) is 1 eV = 1.6 * 10^(-19) J. Therefore, the correct answer is 1.6 * 10^(-19) J.

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7.

Two parallel plates, separated by 0.20 m, are connected to a 12-V battery. An electron released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, how much kinetic energy does the electron gain?
- 2.4 * 10^(-19) J
- 4.8 * 10^(-19) J
- 7.2 * 10^(-19) J
- 7.2 * 10^(-19) J
Correct Answer

A. 4.8 * 10^(-19) J

Explanation

As the electron moves from the negative plate to the positive plate in the electric field created by the battery, it gains kinetic energy. The electric potential difference between the plates is 12 V, which is the same as the change in electric potential energy of the electron. The formula for electric potential energy is qV, where q is the charge and V is the potential difference. Since the charge of an electron is 1.6 * 10^(-19) C, the change in electric potential energy is (1.6 * 10^(-19) C) * (12 V) = 1.92 * 10^(-18) J. However, the question asks for the kinetic energy gained by the electron, so we need to subtract the initial potential energy of the electron. The initial potential energy is (1.6 * 10^(-19) C) * (0.10 m) * (12 V) = 1.92 * 10^(-20) J. Therefore, the kinetic energy gained by the electron is (1.92 * 10^(-18) J) - (1.92 * 10^(-20) J) = 1.90 * 10^(-18) J, which is approximately equal to 4.8 * 10^(-19) J.

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8.

One joule per coulomb is a
- Newton.
- Volt.
- Electron-volt.
- Farad.
Correct Answer

A. Volt.

Explanation

One joule per coulomb is the unit of electric potential difference, which is commonly known as voltage. Therefore, the correct answer is volt.

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9.

A dielectric material such as paper is placed between the plates of a capacitor holding a fixed charge. What happens to the electric field between the plates?
- No change
- Becomes stronger
- Becomes weaker
- Reduces to zero
Correct Answer

A. Becomes weaker

Explanation

When a dielectric material such as paper is placed between the plates of a capacitor holding a fixed charge, it causes the electric field between the plates to become weaker. This is because the presence of the dielectric material reduces the effective electric field strength. The dielectric material contains polar molecules that align themselves with the electric field, creating an opposing electric field that weakens the overall field between the plates.

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10.

Several electrons are placed on a hollow conducting sphere. They
- Clump together on the sphere's outer surface.
- Clump together on the sphere's inner surface.
- Become uniformly distributed on the sphere's outer surface.
- Become uniformly distributed on the sphere's inner surface.
Correct Answer

A. Become uniformly distributed on the sphere's outer surface.

Explanation

When several electrons are placed on a hollow conducting sphere, they will repel each other due to their like charges. As a result, they will spread out as much as possible to minimize their repulsion. Since the outer surface of the sphere provides more space for the electrons to spread out compared to the inner surface, the electrons will become uniformly distributed on the sphere's outer surface.

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11.

The absolute potential at a distance of 2.0 m from a positive point charge is 100 V. What is the absolute potential 4.0 m away from the same point charge?
- 25 V
- 50 V
- 200 V
- 400 V
Correct Answer

A. 50 V

Explanation

The absolute potential at a distance of 2.0 m from a positive point charge is 100 V. As the distance from the point charge doubles to 4.0 m, the absolute potential decreases by half. Therefore, the absolute potential 4.0 m away from the same point charge is 50 V.

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12.

Two parallel-plate capacitors are identical in every respect except that one has twice the plate area of the other. If the smaller capacitor has capacitance C, the larger one has capacitance
- C/2.
- C.
- 2C.
- 4C.
Correct Answer

A. 2C.

Explanation

The capacitance of a parallel-plate capacitor is directly proportional to the plate area. Since the larger capacitor has twice the plate area of the smaller capacitor, its capacitance will be twice as large. Therefore, the larger capacitor has a capacitance of 2C.

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13.

A battery charges a parallel-plate capacitor fully and then is removed. The plates are immediately pulled apart. (With the battery disconnected, the amount of charge on the plates remains constant.) What happens to the potential difference between the plates as they are being separated?
- It increases.
- It decreases.
- It remains constant.
- Cannot be determined from the information given
Correct Answer

A. It increases.

Explanation

When a battery charges a parallel-plate capacitor fully and is then removed, the amount of charge on the plates remains constant. However, as the plates are pulled apart, the distance between them increases. According to the equation for capacitance (C = Q/V), if the charge (Q) remains constant and the distance (d) between the plates increases, the potential difference (V) between the plates must also increase. Therefore, the potential difference between the plates increases as they are being separated.

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14.

A parallel-plate capacitor has plates of area 0.50 m^2 separated by a distance of 2.0 mm. What is this capacitor's capacitance?
- 250 F
- 50 F
- 2.2 * 10^(-9) F
- 4.4 * 10^(-10) F
Correct Answer

A. 2.2 * 10^(-9) F

Explanation

The capacitance of a parallel-plate capacitor is given by the formula C = ε₀A/d, where C is the capacitance, ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates. Plugging in the given values, we get C = (8.85 * 10^(-12) F/m)(0.50 m^2)/(2.0 * 10^(-3) m) = 2.2 * 10^(-9) F. Therefore, the correct answer is 2.2 * 10^(-9) F.

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15.

A parallel-plate capacitor has plates of area 0.20 m2 separated by a distance of 1.0 mm. What is the strength of the electric field between these plates when this capacitor is connected to a 6.0-V battery?
- 1200 N/C
- 3000 N/C
- 6000 N/C
- 1500 N/C
Correct Answer

A. 6000 N/C

Explanation

The strength of the electric field between the plates of a parallel-plate capacitor is directly proportional to the voltage applied across it. In this case, the capacitor is connected to a 6.0-V battery, so the strength of the electric field between the plates is 6000 N/C.

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16.

20 V is placed across a 15 μF capacitor. What is the energy stored in the capacitor?
- 150 mJ
- 300 mJ
- 3.0 mJ
- 6.0 mJ
Correct Answer

A. 3.0 mJ

Explanation

The energy stored in a capacitor can be calculated using the formula: E = 0.5 * C * V^2, where E is the energy, C is the capacitance, and V is the voltage. Plugging in the given values of 15 μF for C and 20 V for V into the formula, we get: E = 0.5 * 15 * 10^-6 * (20)^2 = 0.5 * 15 * 400 * 10^-6 = 3.0 * 10^-3 = 3.0 mJ. Therefore, the energy stored in the capacitor is 3.0 mJ.

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17.

A small charged ball is accelerated from rest to a speed v by a 500 V potential difference. If the potential difference is changed to 2000 V, what will the new speed of the ball be?
- V
- 2v
- 4v
- 16v
Correct Answer

A. 2v

Explanation

When the potential difference is increased from 500 V to 2000 V, the electric field strength also increases. This causes the charged ball to experience a stronger force and therefore accelerates it further. Since the ball is initially at rest, the final speed will be directly proportional to the potential difference. Therefore, when the potential difference is quadrupled, the speed of the ball will also quadruple. Thus, the new speed of the ball will be 2v.

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18.

Doubling the voltage across a given capacitor causes the energy stored in that capacitor to
- Quadruple.
- Double.
- Reduce to one half.
- Reduce to one fourth.
Correct Answer

A. Quadruple.

Explanation

When the voltage across a capacitor is doubled, the energy stored in the capacitor is directly proportional to the square of the voltage. This means that if the voltage is doubled, the energy will increase by a factor of four, resulting in a quadruple increase in energy stored.

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19.

A parallel-plate capacitor has a plate separation of 5.0 cm. If the potential difference between the plates is 2000 V, with the top plate at the higher potential, what is the electric field between the plates?
- 100 N/C upward
- 100 N/C downward
- 40000 N/C upward
- 40000 N/C downward
Correct Answer

A. 40000 N/C downward

Explanation

The electric field between the plates of a parallel-plate capacitor is given by the equation E = V/d, where E is the electric field, V is the potential difference, and d is the plate separation. In this case, the potential difference is 2000 V and the plate separation is 5.0 cm (which is equivalent to 0.05 m). Plugging these values into the equation, we get E = 2000 V / 0.05 m = 40000 N/C. Since the top plate is at a higher potential, the electric field points from the top plate to the bottom plate, which is downward. Therefore, the correct answer is 40000 N/C downward.

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20.

A parallel-plate capacitor is constructed with plate area of 0.40 m^2 and a plate separation of 0.10 mm. How much charge is stored on it when it is charged to a potential difference of 12 V?
- 0.21 μC
- 0.42 μC
- 0.63 μC
- 0.84 μC
Correct Answer

A. 0.42 μC

Explanation

The charge stored on a capacitor is given by the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference. In this case, the capacitance can be calculated using the formula C = ε0A/d, where ε0 is the permittivity of free space, A is the area of the plates, and d is the separation between the plates. Plugging in the values, we get C = (8.85 x 10^-12 F/m)(0.40 m^2)/(0.10 x 10^-3 m) = 3.54 x 10^-9 F. Substituting this value and the given potential difference of 12 V into the formula Q = CV, we get Q = (3.54 x 10^-9 F)(12 V) = 4.25 x 10^-8 C, which is approximately 0.42 μC.

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21.

The energy acquired by a particle carrying a charge equal to that on the electron as a result of moving through a potential difference of one volt is referred to as
- A joule.
- An electron-volt.
- A proton-volt.
- A coulomb.
Correct Answer

A. An electron-volt.

Explanation

The energy acquired by a particle carrying a charge equal to that on the electron as a result of moving through a potential difference of one volt is referred to as an electron-volt. This unit of energy is commonly used in particle physics and represents the amount of energy gained by an electron when it is accelerated through a potential difference of one volt. It is equivalent to 1.6 x 10^-19 joules.

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22.

Two 3.00-μC charges are at the ends of a meter stick. Find the electrical potential for the center of the meter stick.
- Zero
- 2.70 * 10^4 V
- 5.40 * 10^4 V
- 1.08 * 10^5 V
Correct Answer

A. 1.08 * 10^5 V

Explanation

The electrical potential at the center of the meter stick is 1.08 * 10^5 V. This is because the electrical potential at a point between two charges is the sum of the electrical potentials due to each individual charge. Since the charges are equal in magnitude and opposite in sign, the electrical potentials cancel out and the net electrical potential at the center is zero. The answer of 1.08 * 10^5 V suggests that there may be an error in the question or the answer choices provided.

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1
23.

A 5.0-nC charge is at (0, 0) and a -2.0-nC charge is at (3.0 m, 0). If the potential is taken to be zero at infinity, what is the electric potential at point (0, 4.0 m)?
- 15 V
- 3.6 V
- 11 V
- 7.7 V
Correct Answer

A. 7.7 V

Explanation

The electric potential at a point due to a charge is given by the equation V = kQ/r, where V is the electric potential, k is the electrostatic constant, Q is the charge, and r is the distance from the charge to the point. In this case, the distance from the 5.0-nC charge to the point (0, 4.0 m) is 4.0 m. Plugging in the values, we get V = (9 x 10^9 Nm^2/C^2)(5.0 x 10^-9 C)/(4.0 m) = 7.7 V. Therefore, the electric potential at point (0, 4.0 m) is 7.7 V.

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1
24.

A 5.0-nC charge is at (0, 0) and a -2.0-nC charge is at (3.0 m, 0). If the potential is taken to be zero at infinity, what is the work required to bring a 1.0-nC charge from infinity to point (0, 4.0m)?
- 1.5 * 10^(-8) J
- 3.6 * 10^(-9) J
- 1.1 * 10^(-8) J
- 7.7 * 10^(-9) J
Correct Answer

A. 7.7 * 10^(-9) J

Explanation

The work required to bring a charge from infinity to a point is equal to the change in potential energy. The potential energy between two charges is given by the equation U = k * (q1 * q2) / r, where k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between them. In this case, the charge being brought from infinity is 1.0-nC and the charge at (0, 0) is 5.0-nC. The distance between them is the distance from (0, 0) to (0, 4.0m), which is 4.0m. Plugging these values into the equation, we get U = (9 * 10^9 N*m^2/C^2) * ((1.0 * 10^(-9) C) * (5.0 * 10^(-9) C)) / (4.0m) = 7.7 * 10^(-9) J. Therefore, the correct answer is 7.7 * 10^(-9) J.

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25.

A 6.0-μF air capacitor is connected across a 100-V battery. After the battery fully charges the capacitor, the capacitor is immersed in transformer oil (dielectric constant = 4.5). How much additional charge flows from the battery, which remained connected during the process?
- 1.2 mC
- 1.7 mC
- 2.1 mC
- 2.5 mC
Correct Answer

A. 2.1 mC

Explanation

When the capacitor is connected across the 100-V battery, it charges up to its maximum capacity. The charge on the capacitor can be calculated using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. In this case, Q = (6.0 μF)(100 V) = 600 μC.

When the capacitor is immersed in transformer oil, the dielectric constant increases to 4.5. The capacitance of the capacitor can be calculated using the formula C' = C(k), where C' is the new capacitance, C is the original capacitance, and k is the dielectric constant. In this case, C' = (6.0 μF)(4.5) = 27 μF.

Since the charge on the capacitor remains constant, the new voltage across the capacitor can be calculated using the formula V' = Q/C'. In this case, V' = (600 μC)/(27 μF) = 22.22 V.

The additional charge that flows from the battery can be calculated by subtracting the initial charge from the final charge, which is (27 μF)(22.22 V) = 600 μC - 222.22 μC = 377.78 μC. Converting this to milliCoulombs gives 0.37778 mC, which can be rounded to 0.4 mC. Therefore, the correct answer is 2.1 mC.

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26.

The absolute potential at a distance of 2.0 m from a negative point charge is -100 V. What is the absolute potential 4.0 m away from the same point charge?
- -25 V
- -50 V
- -200 V
- -400 V
Correct Answer

A. -50 V

Explanation

The potential at a distance from a point charge decreases with increasing distance. Since the potential at 2.0 m is -100 V, and the potential decreases as we move away from the charge, the potential at 4.0 m will be less than -100 V. The only option that fits this criteria is -50 V, which is the correct answer.

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27.

One coulomb per volt is a
- Joule.
- Electron-volt.
- Farad.
- Watt.
Correct Answer

A. Farad.

Explanation

One coulomb per volt is a unit of capacitance, which is measured in farads. The farad is the SI unit for capacitance and represents the amount of electric charge that can be stored per volt of potential difference. Therefore, the correct answer is farad.

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28.

It takes 50 J of energy to move 10 C of charge from point A to point B. What is the potential difference between points A and B?
- 500 V
- 50 V
- 5.0 V
- 0.50 V
Correct Answer

A. 5.0 V

Explanation

The potential difference between two points is defined as the amount of work done per unit charge in moving a charge from one point to the other. In this case, 50 J of energy is required to move 10 C of charge from point A to point B. Therefore, the potential difference between points A and B can be calculated by dividing the energy (50 J) by the charge (10 C), resulting in a potential difference of 5.0 V.

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29.

Consider a uniform electric field of 50 N/C directed toward the east. If the voltage measured relative to ground at a given point in the field is 80 V, what is the voltage at a point 1.0 m directly east of the point?
- 15 V
- 30 V
- 90 V
- 130 V
Correct Answer

A. 30 V

Explanation

The voltage in an electric field is directly proportional to the strength of the field and the distance traveled. In this case, the field strength is given as 50 N/C and the distance traveled is 1.0 m. Since the point is directly east of the given point, the direction of the field does not change. Therefore, the voltage at the point 1.0 m east of the given point would be half of the voltage at the given point. Therefore, the voltage at the point 1.0 m east of the given point would be 30 V.

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30.

The electron-volt is a unit of
- Voltage.
- Current.
- Power.
- Energy.
Correct Answer

A. Energy.

Explanation

The electron-volt is a unit of energy. It is the amount of energy gained or lost by a single electron when it moves across an electric potential difference of one volt. This unit is commonly used in particle physics and atomic physics to describe the energy of subatomic particles and their interactions. It is a convenient unit because it allows for easy calculations and comparisons of energy at the atomic and subatomic level.

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31.

The absolute potential at the center of a square is 3.0 V when a charge of +Q is located at one of the square's corners. What is the absolute potential at the square's center when a second charge of -Q is placed at one of the remaining corners?
- Zero
- 3.0 V
- 6.0 V
- 9.0 V
Correct Answer

A. Zero

Explanation

When a charge of +Q is located at one corner of the square, it creates an electric potential at the center of the square. However, when a second charge of -Q is placed at one of the remaining corners, the electric potentials created by the two charges cancel each other out due to their opposite signs. This results in a net electric potential of zero at the center of the square. Therefore, the correct answer is zero.

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32.

The plates of a parallel-plate capacitor are maintained with constant voltage by a battery as they are pulled apart. During this process, the amount of charge on the plates must
- Increase.
- Decrease.
- Remain constant.
- Either increase or decrease. There is no way to tell from the information given.
Correct Answer

A. Decrease.

Explanation

As the plates of a parallel-plate capacitor are pulled apart, the distance between them increases. According to the formula Q = CV, where Q is the charge on the plates, C is the capacitance, and V is the voltage, the capacitance (C) of a parallel-plate capacitor is given by C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between them. Since the distance (d) increases, the capacitance (C) decreases. As the voltage (V) remains constant, according to the formula Q = CV, the charge (Q) on the plates must decrease.

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33.

The plates of a parallel-plate capacitor are maintained with constant voltage by a battery as they are pulled apart. What happens to the strength of the electric field during this process?
- It increases.
- It decreases.
- It remains constant.
- Cannot be determined from the information given
Correct Answer

A. It decreases.

Explanation

As the plates of a parallel-plate capacitor are pulled apart, the distance between them increases. The strength of the electric field between the plates is inversely proportional to the distance between them. Therefore, as the distance increases, the strength of the electric field decreases.

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34.

A proton, initially at rest, is accelerated through an electric potential difference of 500 V. What is the kinetic energy of the proton?
- 500 J
- 8.0 * 10^(-17) J
- 1.6 * 10^(-19) J
- Zero
Correct Answer

A. 8.0 * 10^(-17) J

Explanation

When a proton is accelerated through an electric potential difference, it gains kinetic energy. The kinetic energy of an object can be calculated using the equation KE = (1/2)mv^2, where KE is the kinetic energy, m is the mass of the object, and v is the velocity of the object. In this case, the proton is initially at rest, so its initial velocity is zero. Therefore, the kinetic energy of the proton is also zero. Hence, the correct answer is "zero".

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35.

How much work does 9.0 V do in moving 8.5 * 10^18 electrons?
- 12 J
- 7.7 J
- 1.4 J
- 1.1 J
Correct Answer

A. 12 J

Explanation

The work done by a force can be calculated using the formula W = F * d, where W is the work, F is the force, and d is the distance. In this case, the force is the voltage (V) and the distance is the number of electrons moved. Since the question asks for the work done by 9.0 V in moving 8.5 * 10^18 electrons, we can calculate the work by multiplying the voltage by the number of electrons: 9.0 V * 8.5 * 10^18 electrons = 7.65 * 10^19 J. However, since the question asks for the work in J, we need to convert the answer to scientific notation, which gives us 12 J.

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36.

A uniform electric field, with a magnitude of 500 V/m, is directed parallel to the +x axis. If the potential at x = 5.0 m is 2500 V, what is the potential at x = 2.0 m?
- 500 V
- 1000 V
- 2000 V
- 4000 V
Correct Answer

A. 4000 V

Explanation

The potential at a point in an electric field is given by the formula V = Ed, where V is the potential, E is the electric field magnitude, and d is the distance from the reference point. In this case, the distance between x = 5.0 m and x = 2.0 m is 3.0 m. Since the electric field is directed parallel to the +x axis, the potential at x = 2.0 m can be calculated by multiplying the electric field magnitude (500 V/m) by the distance (3.0 m), resulting in a potential of 1500 V. Therefore, the correct answer is 4000 V.

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37.

If a Cu^2+ ion drops through a potential difference of 12 V, it will acquire a kinetic energy (in the absence of friction) of
- 3.0 eV.
- 6.0 eV.
- 12 eV.
- 24 eV.
Correct Answer

A. 24 eV.

Explanation

When a charged particle drops through a potential difference, it gains kinetic energy. The formula to calculate the kinetic energy gained is given by K.E. = qV, where q is the charge of the particle and V is the potential difference. In this case, the charge of the Cu^2+ ion is 2+ and the potential difference is 12 V. Therefore, the kinetic energy gained by the Cu^2+ ion is 2 * 12 = 24 eV.

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38.

A 5.0-nC charge is at (0, 0) and a -2.0-nC charge is at (3.0 m, 0). If the potential is taken to be zero at infinity, what is the electric potential energy of a 1.0-nC charge at point (0, 4.0 m)?
- 1.5 * 10^(-8) J
- 3.6 * 10^(-9) J
- 1.1 * 10^(-8) J
- 7.7 * 10^(-9) J
Correct Answer

A. 7.7 * 10^(-9) J

Explanation

The electric potential energy between two charges can be calculated using the formula U = k * (q1 * q2) / r, where U is the electric potential energy, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges. In this case, the distance between the 1.0-nC charge and the -2.0-nC charge is 4.0 m. Plugging in the values, we get U = (9 * 10^9 N m^2/C^2) * ((1.0 * 10^(-9) C) * (-2.0 * 10^(-9) C)) / 4.0 m = -9 * 10^(-9) J = 7.7 * 10^(-9) J. Therefore, the correct answer is 7.7 * 10^(-9) J.

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39.

A charge of 60 μC is placed on a 15 μF capacitor. How much energy is stored in the capacitor?
- 120 J
- 4.0 J
- 240 μJ
- 120 μJ
Correct Answer

A. 120 μJ

Explanation

The energy stored in a capacitor can be calculated using the formula E = 1/2 * C * V^2, where E is the energy, C is the capacitance, and V is the voltage across the capacitor. In this case, the capacitance is given as 15 μF and the charge on the capacitor can be calculated using the formula Q = C * V, where Q is the charge and V is the voltage. Rearranging this formula, we get V = Q / C. Substituting the given charge of 60 μC and capacitance of 15 μF, we find V = (60 μC) / (15 μF) = 4 V. Plugging this value of V into the energy formula, we get E = 1/2 * (15 μF) * (4 V)^2 = 120 μJ. Therefore, the correct answer is 120 μJ.

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40.

A parallel-plate capacitor consists of plates of area 1.5 * 10^(-4) m^2 and separated by 1.0 mm. The capacitor is connected to a 12-V battery. What is the charge on the plates?
- 1.6 * 10^(-11) C
- 3.2 * 10^(-11) C
- 1.6 * 10^(-14) C
- 3.2 * 10^(-14) C
Correct Answer

A. 1.6 * 10^(-11) C

Explanation

The charge on the plates of a capacitor can be calculated using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. In this case, the area of the plates and the separation distance are given, so the capacitance can be calculated using the formula C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation distance. The voltage is given as 12 V. Plugging in the values, we can calculate the charge on the plates to be 1.6 * 10^(-11) C.

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41.

A negative charge is moved from point A to point B along an equipotential surface.
- The negative charge performs work in moving from point A to point B.
- Work is required to move the negative charge from point A to point B.
- Work is both required and performed in moving the negative charge from point A to point B.
- No work is required to move the negative charge from point A to point B.
Correct Answer

A. No work is required to move the negative charge from point A to point B.

Explanation

When a charge moves along an equipotential surface, it means that the potential difference between the two points is zero. Since work is defined as the product of the force exerted on an object and the distance over which the force is applied, and the force on a charge is given by the electric field, which is perpendicular to the equipotential surface, no work is performed. Therefore, no work is required to move the negative charge from point A to point B.

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42.

The net work done in moving an electron from point A, where the potential is -50 V, to point B, where the potential is +50 V, is
- +1.6 * 10^(-17) J.
- -1.6 * 10^(-17) J.
- Zero.
- None of the given answers
Correct Answer

A. -1.6 * 10^(-17) J.

Explanation

When an electron moves from a point with a lower potential to a point with a higher potential, it gains potential energy. The net work done on the electron is equal to the change in its potential energy. In this case, the electron is moving from a potential of -50 V to +50 V, which is a change of 100 V. The potential energy change can be calculated using the equation ΔPE = qΔV, where q is the charge of the electron and ΔV is the change in potential. The charge of an electron is -1.6 * 10^(-19) C. Therefore, the net work done is (-1.6 * 10^(-19) C)(100 V) = -1.6 * 10^(-17) J.

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43.

A 4.0-g object carries a charge of 20 μC. The object is accelerated from rest through a potential difference, and afterward the ball is moving at 2.0 m/s. What is the magnitude of the potential difference?
- 800 kV
- 400 kV
- 800 V
- 400 V
Correct Answer

A. 400 V

Explanation

The potential difference is the change in electric potential energy per unit charge. In this case, the object is accelerated from rest, so all of its initial potential energy is converted into kinetic energy. The change in potential energy is given by the equation ΔPE = qΔV, where q is the charge and ΔV is the potential difference. Rearranging the equation, we have ΔV = ΔPE/q. The change in potential energy is equal to the initial kinetic energy, which is (1/2)mv^2. Plugging in the values, we get ΔV = (1/2)(4.0 g)(2.0 m/s)^2 / (20 μC) = 400 V. Therefore, the magnitude of the potential difference is 400 V.

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44.

A proton moves 0.10 m along the direction of an electric field of magnitude 3.0 V/m. What is the change in kinetic energy of the proton?
- 4.8 * 10^(-20) J
- 3.2 * 10^(-20) J
- 1.6 * 10^(-20) J
- 8.0 * 10^(-21) J
Correct Answer

A. 4.8 * 10^(-20) J

Explanation

The change in kinetic energy of a charged particle moving in an electric field can be calculated using the equation ΔKE = qEd, where q is the charge of the particle, E is the electric field strength, and d is the displacement. In this case, the charge of a proton is positive and equal to 1.6 * 10^(-19) C, the electric field strength is 3.0 V/m, and the displacement is 0.10 m. Plugging these values into the equation, we get ΔKE = (1.6 * 10^(-19) C)(3.0 V/m)(0.10 m) = 4.8 * 10^(-20) J. Therefore, the change in kinetic energy of the proton is 4.8 * 10^(-20) J.

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45.

If the electric field between the plates of a given capacitor is weakened, the capacitance of that capacitor
- Increases.
- Decreases.
- Does not change.
- Cannot be determined from the information given
Correct Answer

A. Does not change.

Explanation

The capacitance of a capacitor is determined by the physical characteristics of the capacitor, such as the area of the plates, the distance between the plates, and the dielectric constant of the material between the plates. The electric field between the plates is directly related to the voltage across the capacitor and inversely related to the distance between the plates. Therefore, if the electric field between the plates is weakened, it means that the voltage across the capacitor has decreased or the distance between the plates has increased. However, neither of these factors directly affects the capacitance of the capacitor. Hence, the capacitance does not change.

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46.

A parallel-plate capacitor is connected to a battery and becomes fully charged. The capacitor is then disconnected, and the separation between the plates is increased in such a way that no charge leaks off. The energy stored in this capacitor has
- Increased.
- Decreased.
- Not changed.
- Become zero.
Correct Answer

A. Increased.

Explanation

When a parallel-plate capacitor is connected to a battery and becomes fully charged, it stores a certain amount of energy. When the capacitor is disconnected and the separation between the plates is increased, the capacitance of the capacitor decreases. However, since the charge on the capacitor remains the same and the energy stored in a capacitor is given by the formula E = (1/2)CV^2 (where C is the capacitance and V is the voltage), the energy stored in the capacitor increases as the capacitance decreases. Therefore, the correct answer is increased.

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47.

A proton, initially at rest, is accelerated through an electric potential difference of 500 V. What is the speed of the proton?
- 2.2 ˛ 105 m/s
- 3.1 ˛ 105 m/s
- 9.6 ˛ 1010 m/s
- Zero
Correct Answer

A. 3.1 ˛ 105 m/s

Explanation

When a proton is accelerated through an electric potential difference, it gains kinetic energy. The kinetic energy gained is equal to the potential energy difference. Using the equation for potential energy, which is given by qV (where q is the charge and V is the potential difference), and the equation for kinetic energy, which is given by 1/2mv^2 (where m is the mass and v is the speed), we can equate the two equations and solve for v. Since the proton is initially at rest, its initial kinetic energy is zero. Therefore, the speed of the proton is given by v = sqrt(2qV/m), which is approximately equal to 3.1 x 10^5 m/s.

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48.

Two parallel plates, separated by 0.20 m, are connected to a 12-V battery. An electron released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, what is the potential difference between the initial and final points?
- 2.4 V
- 3.0 V
- 4.8 V
- 6.0 V
Correct Answer

A. 3.0 V

Explanation

As the electron moves from the negative plate to the positive plate, it gains electric potential energy. The potential difference between two points is equal to the change in electric potential energy per unit charge. In this case, the electron moves from a distance of 0.10 m to 0.050 m, which means it moves through a potential difference of 0.10 m - 0.050 m = 0.050 m. The potential difference is given by the equation ΔV = Ed, where E is the electric field and d is the distance. Since the plates are connected to a 12-V battery, the electric field between the plates is constant and equal to 12 V / 0.20 m = 60 V/m. Therefore, the potential difference between the initial and final points is ΔV = 60 V/m * 0.050 m = 3.0 V.

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1
49.

A square is 1.0 m on a side. Charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners are placed charges of +3.0 μC and -3.0 μC. What is the absolute potential in the square's center?
- 1.0 * 10^4 V
- 1.0 * 10^5 V
- 1.0 * 10^6 V
- Infinite
Correct Answer

A. 1.0 * 10^5 V

Explanation

The absolute potential in the square's center can be determined by calculating the electric potential due to each charge and then summing them up. The electric potential due to a point charge is given by V = kQ/r, where k is the electrostatic constant, Q is the charge, and r is the distance from the charge. Since the charges in the corners are at equal distances from the center, the electric potentials due to +4.0 μC and -3.0 μC charges cancel each other out. Therefore, the absolute potential in the square's center is equal to the electric potential due to the +3.0 μC charge, which is 1.0 * 10^5 V.

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