Physics Test: Important Practice Questions On Electrostatics!

1. Four charges +q, +q, -q and -q respectively are placed at the corners A,B,C and D of a square of side a. The electric potential at the centre O of the square is

Q/4πεoa

2q/4πεoa

4q/4πεoa

Zero

The electric potential at the center O of the square is zero because the positive and negative charges cancel each other out. The two positive charges create a positive potential, while the two negative charges create a negative potential. Since the magnitudes of the charges are equal, their potentials cancel each other out, resulting in a net potential of zero at the center.

Explanation

The electric potential at the center O of the square is zero because the positive and negative charges cancel each other out. The two positive charges create a positive potential, while the two negative charges create a negative potential. Since the magnitudes of the charges are equal, their potentials cancel each other out, resulting in a net potential of zero at the center.

2. The work done in moving 500 μC charge between two points on equipotential surface is

Zero

Finite positive

Finite negative

Infinite

When a charge is moved between two points on an equipotential surface, the potential difference between those points is zero. This means that the electric field and the force experienced by the charge are also zero. Since work done is defined as the product of force and displacement, and the force is zero, the work done in moving the charge is also zero. Therefore, the correct answer is zero.

Explanation

When a charge is moved between two points on an equipotential surface, the potential difference between those points is zero. This means that the electric field and the force experienced by the charge are also zero. Since work done is defined as the product of force and displacement, and the force is zero, the work done in moving the charge is also zero. Therefore, the correct answer is zero.

3. The electrostatic force between two point charges kept at a distance d apart , in a medium εr =6, is 0.3N. The force between them at the same separation in vacuum is

20 N

0.5 N

1.8 N

2 N

The electrostatic force between two point 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. The formula for Coulomb's law in a medium is F = (1/4πεr) * (q1 * q2) / r^2, where εr is the relative permittivity of the medium. In this case, the force in the medium is 0.3 N. When the charges are in a vacuum, the relative permittivity is 1, so the force in vacuum can be calculated using the same formula with εr = 1. Thus, the force in vacuum is (1/4π) * (q1 * q2) / r^2. Since the charges and distance are the same, the force in vacuum would be (1/6) * 0.3 N = 0.05 N. Therefore, the correct answer is 1.8 N.

Explanation

The electrostatic force between two point 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. The formula for Coulomb's law in a medium is F = (1/4πεr) * (q1 * q2) / r^2, where εr is the relative permittivity of the medium. In this case, the force in the medium is 0.3 N. When the charges are in a vacuum, the relative permittivity is 1, so the force in vacuum can be calculated using the same formula with εr = 1. Thus, the force in vacuum is (1/4π) * (q1 * q2) / r^2. Since the charges and distance are the same, the force in vacuum would be (1/6) * 0.3 N = 0.05 N. Therefore, the correct answer is 1.8 N.

4. A hollow metal ball carrying an electric charge produces no electric field at points

Outside the sphere

On its surface

Inside the sphere

At a distance more than twice

Inside the sphere is the correct answer because the electric field inside a hollow metal ball carrying an electric charge is zero. This is due to the fact that the charges on the inner surface of the hollow ball redistribute themselves in such a way that the electric field they produce cancels out. As a result, there is no net electric field inside the sphere.

Explanation

Inside the sphere is the correct answer because the electric field inside a hollow metal ball carrying an electric charge is zero. This is due to the fact that the charges on the inner surface of the hollow ball redistribute themselves in such a way that the electric field they produce cancels out. As a result, there is no net electric field inside the sphere.

5. Which of the following quantities is scalar?

Dipole moment

Electric force

Electric field

Electric potential

Electric potential is a scalar quantity. Scalar quantities have magnitude but no direction. Electric potential represents the amount of electric potential energy per unit charge at a given point in an electric field. It is a scalar because it only has magnitude and does not have a specific direction associated with it. In contrast, dipole moment, electric force, and electric field are vector quantities because they have both magnitude and direction.

Explanation

Electric potential is a scalar quantity. Scalar quantities have magnitude but no direction. Electric potential represents the amount of electric potential energy per unit charge at a given point in an electric field. It is a scalar because it only has magnitude and does not have a specific direction associated with it. In contrast, dipole moment, electric force, and electric field are vector quantities because they have both magnitude and direction.

6. If a point lies at a distance x from the midpoint of the dipole, the electric potential at this point is proportional to

1/x2

1/x3

1/x4

1/x3/2

The electric potential at a point due to a dipole is inversely proportional to the square of the distance from the midpoint of the dipole. This is because the electric field produced by a dipole decreases with the square of the distance. Therefore, the correct answer is 1/x^2, where x is the distance from the midpoint of the dipole.

Explanation

The electric potential at a point due to a dipole is inversely proportional to the square of the distance from the midpoint of the dipole. This is because the electric field produced by a dipole decreases with the square of the distance. Therefore, the correct answer is 1/x^2, where x is the distance from the midpoint of the dipole.

7. The electric field outside the plates of two oppositely charged plane sheets of charge density σ is

+σ/2εσ

-σ/2εσ

σ/εo

Zero

The electric field outside the plates of two oppositely charged plane sheets of charge density σ is zero. This is because the electric fields created by the two sheets cancel each other out. The positive sheet creates an electric field pointing away from it, while the negative sheet creates an electric field pointing towards it. These two fields have equal magnitudes and opposite directions, resulting in a net electric field of zero outside the plates.

Explanation

The electric field outside the plates of two oppositely charged plane sheets of charge density σ is zero. This is because the electric fields created by the two sheets cancel each other out. The positive sheet creates an electric field pointing away from it, while the negative sheet creates an electric field pointing towards it. These two fields have equal magnitudes and opposite directions, resulting in a net electric field of zero outside the plates.

8. The capacitance of a parallel plate capacitor increases from 5μf to 60μf when a dielectric is filled between the plates.The dielectric constant of the dielectric is

65

55

12

10

When a dielectric is filled between the plates of a parallel plate capacitor, the capacitance increases. The relationship between the capacitance and the dielectric constant is given by the formula C = k * Co, where C is the capacitance with the dielectric, Co is the capacitance without the dielectric, and k is the dielectric constant. In this case, the initial capacitance without the dielectric is 5μF and the final capacitance with the dielectric is 60μF. By substituting these values into the formula, we can solve for k. Therefore, the dielectric constant of the dielectric is 12.

Explanation

When a dielectric is filled between the plates of a parallel plate capacitor, the capacitance increases. The relationship between the capacitance and the dielectric constant is given by the formula C = k * Co, where C is the capacitance with the dielectric, Co is the capacitance without the dielectric, and k is the dielectric constant. In this case, the initial capacitance without the dielectric is 5μF and the final capacitance with the dielectric is 60μF. By substituting these values into the formula, we can solve for k. Therefore, the dielectric constant of the dielectric is 12.

9. The unit of permittivity is

C2N-1m-2

Nm2C-2

Hm-1

NC-2m-2

The unit of permittivity is C2N-1m-2. This unit represents the ability of a material to store electrical energy in an electric field. The unit is derived from the formula for permittivity, which is ε = Q/(EA), where Q is charge, E is electric field, and A is area. The unit C2N-1m-2 represents the ratio of charge squared to force, per unit area and per unit electric field.

Explanation

The unit of permittivity is C2N-1m-2. This unit represents the ability of a material to store electrical energy in an electric field. The unit is derived from the formula for permittivity, which is ε = Q/(EA), where Q is charge, E is electric field, and A is area. The unit C2N-1m-2 represents the ratio of charge squared to force, per unit area and per unit electric field.

10. Electric field intensity is 400Vm^-1 at a distance of 2m from a point charge.It will be 100Vm^-1 at a distance?

50 cm

4 cm

4 m

1.5 m

The electric field intensity decreases with increasing distance from a point charge. In this case, the electric field intensity is 400 Vm-1 at a distance of 2m. Since the field intensity decreases with distance, it will be 100 Vm-1 at a greater distance. The only option that represents a greater distance than 2m is 4m, so the correct answer is 4m.

Explanation

The electric field intensity decreases with increasing distance from a point charge. In this case, the electric field intensity is 400 Vm-1 at a distance of 2m. Since the field intensity decreases with distance, it will be 100 Vm-1 at a greater distance. The only option that represents a greater distance than 2m is 4m, so the correct answer is 4m.

11. Two point charges +4q and +q are placed 30cm apart. At what point on the line joining them the electric field is zero?

15 cm from the charge q

7.5 cm from the charge q

20 cm from the charge 4q

5 cm from the charge q

The electric field is zero at a point on the line joining the charges when the magnitudes of the electric fields due to the two charges cancel each other out. Since the electric field due to a point charge decreases with distance, the electric field due to the charge +4q will be stronger closer to it. Therefore, the electric field due to the charge +q must be stronger at a point closer to it. To cancel out the electric field due to +4q, the point must be farther away from it. Therefore, the electric field is zero at a point 20 cm from the charge +4q.

Explanation

The electric field is zero at a point on the line joining the charges when the magnitudes of the electric fields due to the two charges cancel each other out. Since the electric field due to a point charge decreases with distance, the electric field due to the charge +4q will be stronger closer to it. Therefore, the electric field due to the charge +q must be stronger at a point closer to it. To cancel out the electric field due to +4q, the point must be farther away from it. Therefore, the electric field is zero at a point 20 cm from the charge +4q.

12. Electric potential energy (U) of two point charges is

Q1q2/4πεor2

Q1q2/4πεor

PE cosθ

PE sinθ

The given formula for electric potential energy (U) is correct and is derived from the equation U = k * q1 * q2 / r, where k is the electrostatic constant (1 / 4πεo). The formula q1q2/4πεor represents the potential energy between two point charges q1 and q2 separated by a distance r.

Explanation

The given formula for electric potential energy (U) is correct and is derived from the equation U = k * q1 * q2 / r, where k is the electrostatic constant (1 / 4πεo). The formula q1q2/4πεor represents the potential energy between two point charges q1 and q2 separated by a distance r.

13. A dipole is placed in a uniform electric field with its axis parallel to the field. It experiences

Only a net force

Only a torque

Both a net force and torque

Neither a net force nor a torque

When a dipole is placed in a uniform electric field with its axis parallel to the field, the positive and negative charges of the dipole experience equal and opposite forces. As a result, the net force on the dipole is zero. Additionally, since the forces acting on the charges are equal in magnitude and opposite in direction, they create a torque that cancels out. Therefore, the dipole does not experience a net force or torque in this scenario.

Explanation

When a dipole is placed in a uniform electric field with its axis parallel to the field, the positive and negative charges of the dipole experience equal and opposite forces. As a result, the net force on the dipole is zero. Additionally, since the forces acting on the charges are equal in magnitude and opposite in direction, they create a torque that cancels out. Therefore, the dipole does not experience a net force or torque in this scenario.

14. A glass rod rubbed with silk acquires a charge of +8 x 10^-12C. The number of electrons it has gained or lost

5 x 10-7(gained)

5 x 107(lost)

2 x 10-8(lost)

-8 x 10-12(lost)

When a glass rod is rubbed with silk, it acquires a positive charge. This means that it has lost electrons, as electrons have a negative charge. The given answer, "5 x 10^7 (lost)", indicates that the glass rod has lost 5 x 10^7 electrons, resulting in a positive charge.

Explanation

When a glass rod is rubbed with silk, it acquires a positive charge. This means that it has lost electrons, as electrons have a negative charge. The given answer, "5 x 10^7 (lost)", indicates that the glass rod has lost 5 x 10^7 electrons, resulting in a positive charge.

15. The number of electric lines of force originating from a charge of 1C is

1.129 x 1011

1.6 x 10-19

6.25 x 1018

8.85 x 1012

The number of electric lines of force originating from a charge of 1C is given by Coulomb's law, which states that the electric field due to a point charge is directly proportional to the charge and inversely proportional to the square of the distance from the charge. Therefore, the number of electric lines of force is determined by the magnitude of the charge and the distance at which we are measuring it. The answer 1.129 x 1011 is a possible value for the number of lines of force, but without additional information about the distance, it is not possible to determine the exact value.

Explanation

The number of electric lines of force originating from a charge of 1C is given by Coulomb's law, which states that the electric field due to a point charge is directly proportional to the charge and inversely proportional to the square of the distance from the charge. Therefore, the number of electric lines of force is determined by the magnitude of the charge and the distance at which we are measuring it. The answer 1.129 x 1011 is a possible value for the number of lines of force, but without additional information about the distance, it is not possible to determine the exact value.