12 - Physics - Unit 1 - Electro Statics

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

Zero

Finite positive

Finite negative

Infinite

The work done in moving a charge between two points on an equipotential surface is zero because the potential difference between any two points on an equipotential surface is zero. This means that the electric field is perpendicular to the equipotential surface, and therefore no work is done in moving the charge along that surface.

Explanation

The work done in moving a charge between two points on an equipotential surface is zero because the potential difference between any two points on an equipotential surface is zero. This means that the electric field is perpendicular to the equipotential surface, and therefore no work is done in moving the charge along that surface.

2. 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 center O of the square is (1)

(2)

(3)

(4) zero

(4)

(3)

(2)

(1)

The electric potential at the center O of the square is zero because the charges at corners A and C are equal in magnitude but opposite in sign, canceling out each other's electric potential. The same is true for the charges at corners B and D. Therefore, the net electric potential at the center O is zero.

Explanation

The electric potential at the center O of the square is zero because the charges at corners A and C are equal in magnitude but opposite in sign, canceling out each other's electric potential. The same is true for the charges at corners B and D. Therefore, the net electric potential at the center O is zero.

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

Inside the sphere

On its surface

Outside the sphere

At a distance more than twice

A hollow metal ball carrying an electric charge produces no electric field at points inside the sphere because the electric field inside a conductor in electrostatic equilibrium is zero. The charges on the inner surface of the hollow sphere redistribute themselves in such a way that the electric field due to them cancels out, resulting in no net electric field inside the sphere.

Explanation

A hollow metal ball carrying an electric charge produces no electric field at points inside the sphere because the electric field inside a conductor in electrostatic equilibrium is zero. The charges on the inner surface of the hollow sphere redistribute themselves in such a way that the electric field due to them cancels out, resulting in no net electric field inside the sphere.

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

1.8 N

0.5 N

20 N

2 N

The electrostatic force between two point charges is given by the equation F = (1/4πε) * (q1*q2)/r^2, where F is the force, ε is the permittivity of the medium, q1 and q2 are the charges, and r is the distance between them. In this question, the force is given as 0.3 N in a medium with εr = 6. To find the force in vacuum, we can use the fact that εr = ε/ε0, where ε0 is the permittivity of vacuum. Rearranging the equation, we get F_vacuum = (ε0/εr) * F = (1/6) * 0.3 N = 0.05 N. Therefore, the force between the charges at the same separation in vacuum is 0.05 N, which is closest to 0.5 N.

Explanation

The electrostatic force between two point charges is given by the equation F = (1/4πε) * (q1*q2)/r^2, where F is the force, ε is the permittivity of the medium, q1 and q2 are the charges, and r is the distance between them. In this question, the force is given as 0.3 N in a medium with εr = 6. To find the force in vacuum, we can use the fact that εr = ε/ε0, where ε0 is the permittivity of vacuum. Rearranging the equation, we get F_vacuum = (ε0/εr) * F = (1/6) * 0.3 N = 0.05 N. Therefore, the force between the charges at the same separation in vacuum is 0.05 N, which is closest to 0.5 N.

5. Which of the following quantities is scalar?

Electric potential

Electric force

Electric field

Dipole moment

Electric potential is a scalar quantity because it only has magnitude and no direction. It represents the amount of electric potential energy per unit charge at a given point in an electric field. Unlike electric force, electric field, and dipole moment, electric potential does not have a specific direction associated with it. Therefore, it is considered a scalar quantity.

Explanation

Electric potential is a scalar quantity because it only has magnitude and no direction. It represents the amount of electric potential energy per unit charge at a given point in an electric field. Unlike electric force, electric field, and dipole moment, electric potential does not have a specific direction associated with it. Therefore, it is considered a scalar quantity.

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

Neither a net force nor a torque

Only a torque

Both a net force and torque

Only a net force

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. These forces cancel each other out, resulting in no net force on the dipole. Additionally, since the forces on the positive and negative charges are equal in magnitude and opposite in direction, they also create equal and opposite torques. These torques also cancel each other out, resulting in no net torque on the dipole. Therefore, the dipole experiences neither a net force nor a torque in this situation.

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. These forces cancel each other out, resulting in no net force on the dipole. Additionally, since the forces on the positive and negative charges are equal in magnitude and opposite in direction, they also create equal and opposite torques. These torques also cancel each other out, resulting in no net torque on the dipole. Therefore, the dipole experiences neither a net force nor a torque in this situation.

7. If a point lies at a distance x from the midpoint of the dipole, the electric potential at this point is proportional to (1)

(2)

(3)

(4)

(1)

(2)

(3)

(4)

The electric potential at a point due to a dipole is inversely proportional to the distance from the midpoint of the dipole. Therefore, if a point lies at a distance x from the midpoint of the dipole, the electric potential at this point is proportional to 1/x. Hence, the correct answer is (1).

Explanation

The electric potential at a point due to a dipole is inversely proportional to the distance from the midpoint of the dipole. Therefore, if a point lies at a distance x from the midpoint of the dipole, the electric potential at this point is proportional to 1/x. Hence, the correct answer is (1).

8. The electric field outside the plates of two oppositely charged plane sheets of charge density σ is (1)

(2)

(3)

(4) zero

(4)

(3)

(2)

(1)

The electric field outside the plates of two oppositely charged plane sheets of charge density σ is zero. This is because the electric field created by one sheet cancels out the electric field created by the other sheet due to their opposite charges. Therefore, outside the plates, the net electric field is zero.

Explanation

The electric field outside the plates of two oppositely charged plane sheets of charge density σ is zero. This is because the electric field created by one sheet cancels out the electric field created by the other sheet due to their opposite charges. Therefore, outside the plates, the net electric field is zero.

9. The unit of permittivity is (1) C² N⁻¹ m⁻² (2) N m² C⁻² (3) H m⁻¹ (4) N C⁻² m⁻²

(1)

(2)

(3)

(4)

The unit of permittivity is given by C2 N−1 m−2. This unit represents the amount of electric charge (Coulombs) per unit area (m2) required to produce a unit electric field (N). Therefore, option (1) is the correct answer.

Explanation

The unit of permittivity is given by C2 N−1 m−2. This unit represents the amount of electric charge (Coulombs) per unit area (m2) required to produce a unit electric field (N). Therefore, option (1) is the correct answer.

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

12

55

65

10

not-available-via-ai

Explanation

not-available-via-ai

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

20 cm from the charge 4q

15 cm from the charge q

7.5 cm from the charge q

5 cm from the charge q

The electric field at a point on the line joining two charges is zero when the magnitudes of the electric fields produced by the charges at that point are equal and opposite. In this case, the charge +4q produces an electric field that is four times stronger than the electric field produced by the charge +q. To balance out the stronger electric field, the point where the electric field is zero must be closer to the charge +4q. Therefore, the correct answer is 20 cm from the charge 4q.

Explanation

The electric field at a point on the line joining two charges is zero when the magnitudes of the electric fields produced by the charges at that point are equal and opposite. In this case, the charge +4q produces an electric field that is four times stronger than the electric field produced by the charge +q. To balance out the stronger electric field, the point where the electric field is zero must be closer to the charge +4q. Therefore, the correct answer is 20 cm from the charge 4q.

12. Elecric field intensity is 400 V m⁻¹ at a distance of 2 m from a point charge. It will be 100 V m⁻¹ at a distance?

4 m

4 cm

1.5 m

50 cm

The electric field intensity decreases with distance from a point charge. Since the electric field intensity is 400 V m−1 at a distance of 2 m, it will be 100 V m−1 at a distance that is 4 times greater, which is 4 m.

Explanation

The electric field intensity decreases with distance from a point charge. Since the electric field intensity is 400 V m−1 at a distance of 2 m, it will be 100 V m−1 at a distance that is 4 times greater, which is 4 m.

13. The number of electric lines of force originating from a charge of 1 C is (1) (2) (3) (4)

(1)

(2)

(3)

(4)

The number of electric lines of force originating from a charge of 1 C is directly proportional to the magnitude of the charge. Since the charge is given as 1 C, there will be a single electric line of force originating from it.

Explanation

The number of electric lines of force originating from a charge of 1 C is directly proportional to the magnitude of the charge. Since the charge is given as 1 C, there will be a single electric line of force originating from it.

14. Electric potential energy (U) of two point charges is (1)

(2)

(3) pE cos θ (4) pE sin θ

(2)

(3)

(1)

(4)

The correct answer is (2) because the electric potential energy (U) of two point charges is given by the formula U = k(q1*q2)/r, where k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the charges. The given option (2) matches this formula, indicating that it is the correct answer.

Explanation

The correct answer is (2) because the electric potential energy (U) of two point charges is given by the formula U = k(q1*q2)/r, where k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the charges. The given option (2) matches this formula, indicating that it is the correct answer.

15. A glass rod rubbed with silk acquires a charge of . The number of electrons it has gained or lost (1) (gained) (2) (lost) (3) (lost) (4) (lost)

(2)

(1)

(4)

(3)

When a glass rod is rubbed with silk, it acquires a positive charge. This means that it has lost electrons, as electrons are negatively charged particles. Therefore, the correct answer is (2) lost.

Explanation

When a glass rod is rubbed with silk, it acquires a positive charge. This means that it has lost electrons, as electrons are negatively charged particles. Therefore, the correct answer is (2) lost.