Electric Potential - NEET Physics Questions
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Electric Potential

Question 11: easy

Sixty four identical drops of water having equal charge combine to form a bigger drop. The factor by which potential of bigger drop change in comparison to a small drop is

1. 64
2. 32
3. 16
4. 8
View Answer

When \(N = 64\) drops combine, the new radius is \(R = N^{1/3}r = 4r\). The new charge is \(Q = Nq = 64q\). The potential of the bigger drop is \(V' = \frac{kQ}{R} = \frac{k(64q)}{4r} = 16 \left(\frac{kq}{r}\right) = 16V\).

Question 12: easy

Sixty four identical drops of water having equal charge combine to form a bigger drop. The factor by which potential of bigger drop change in comparison to a small drop is

1. 64
2. 32
3. 16
4. 8
View Answer

By conserving volume, \(R = n^{1/3}r = 64^{1/3}r = 4r\). The total charge is \(Q = 64q\). Thus, potential of the big drop is \(V' = \frac{kQ}{R} = \frac{64kq}{4r} = 16V\).

Question 13: easy

Assertion (A): The whole charge of a conductor cannot be transferred to another conductor.


Reason (R): The total transfer of charge from one to another is not possible.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Both (A) and (R) are false. Charge can be completely transferred between conductors, for example, by induction or conduction, if placed inside a hollow conductor and connected.

Question 14: easy

Assertion (A): Electrostatic field inside a conducting shell is always zero.


Reason (R): The electrostatic potential is always same from center to surface of a conducting shell.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Concept: Properties of conductors in electrostatic equilibrium.
Formula: \( \vec{E} = -\nabla V \).
Solution: In a conductor, free charges redistribute to make \( \vec{E} = 0 \) inside (A is true). If \( \vec{E} = 0 \) inside, then \( V \) must be constant (R is true) and thus (R) correctly explains (A).

Question 15: easy

The electrostatic potential on the surface of a charged solid conducting sphere is 100 volts. Two statements are made in this regard :


Assertion (A): At any point inside the sphere, electrostatic potential is 100 volt.


Reason (R): At any point inside the sphere, electric field is zero.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Concept: Properties of charged conductors.
Principle: Inside a conductor, \( \vec{E} = 0 \) and \( V = \text{constant} \).
Solution: For a solid conducting sphere, potential is uniform throughout its volume and equals surface potential (A is true). This is because the electric field inside a conductor is zero (R is true), implying constant potential. Thus, (R) correctly explains (A).

Question 16: easy

Assertion (A): If electric field in x-y plane is given by \( \vec{E} = y\hat{i} + x\hat{j} \) then equipotential curve is given by \( xy = \text{ constant} \).


Reason (R): Electric field may not be perpendicular to equipotential surface/curve/line.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Concept: Relation between electric field and equipotential surfaces.
Formula: \( \vec{E} = -\nabla V \).
Solution: From \( \vec{E} = -\nabla V \) for \( \vec{E} = y\hat{i} + x\hat{j} \), potential is \( V = -xy + C \), so equipotential curves are \( xy = \text{ constant} \) (A is true). Electric field lines are always perpendicular to equipotential surfaces (R is false).

Question 17: easy

Assertion (A): When an isolated charged body is connected to earth, all its charge flows to earth and it becomes electrically neutral.


Reason (R): Electric potential of earth is non zero, so the body connected to earth should also attain zero potential.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is true as charges flow to earth. Reason (R) is false because earth's potential is considered zero.

Question 18: easy

Assertion (A): Potential difference between two points in space is zero if electric field at all points in space is zero.


Reason (R): Electric field E at a point P is zero if potential at that point is zero.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Concept: Relation between electric field and potential.
Formula: \( \Delta V = -\int \vec{E} \cdot d\vec{l} \). If \( \vec{E} = 0 \), then \( \Delta V = 0 \).
Solution: If \( \vec{E} = 0 \) everywhere, then potential is constant, so \( \Delta V = 0 \) (A is true). If \( V=0 \) at a point, \( \vec{E} \) is not necessarily zero (e.g., center of a dipole) (R is false).

Question 19: easy

Assertion (A): When an isolated charged body is connected to earth, all its charge flows to earth and it becomes electrically neutral.


Reason (R): Electric potential of earth is non zero, so the body connected to earth should also attain zero potential.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Earth is considered to have a zero potential. When a charged body is connected to Earth, charges flow until the body also attains zero potential, making it electrically neutral. Thus (A) is true and (R) is false.

Question 20: easy

Assertion (A): Potential difference between two points in space is zero if electric field at all points in space is zero.


Reason (R): Electric field \( \vec{E} \) at a point \( P \) is zero if potential at that point is zero.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

If \( \vec{E} = \vec{0} \) everywhere, then \( \vec{E} = -\nabla V = \vec{0} \), implying \( V \) is constant, so \( \Delta V = 0 \). However, \( V = 0 \) at a point does not imply \( \vec{E} = \vec{0} \) (e.g., at the center of an electric dipole). Thus (A) is true and (R) is false.