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.
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.
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
In electrostatic equilibrium, the electric field inside a conductor is zero \( (\vec{E} = \vec{0}) \). Since \( \vec{E} = -\nabla V \), if \( \vec{E} = \vec{0} \), the potential \( V \) must be constant throughout the conductor, from center to surface. Thus (A) and (R) are true, and (R) explains (A).
The electrostatic potential on the surface of a charged solid conducting sphere is \( 100 \text{ volts} \). Two statements are made in this regard :
Assertion (A): At any point inside the sphere, electrostatic potential is \( 100 \text{ volt} \).
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
For a charged solid conducting sphere, the electric field inside is zero \( (\vec{E} = \vec{0}) \). Consequently, the potential \( V \) inside is constant and equal to the potential on its surface. Therefore, both (A) and (R) are true, and (R) correctly explains (A).
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
For \( \vec{E} = y \hat{i} + x \hat{j} \), we have \( dV = -\vec{E} \cdot d\vec{l} = -(y dx + x dy) = -d(xy) \). Integrating, \( V = -xy + C \). Thus, equipotential lines are \( xy = \text{constant} \). Electric field lines are always perpendicular to equipotential surfaces. Thus (A) is true and (R) is false.
Assertion (A): Distance of closest approach for free target is more than that for fixed target.
Reason (R): Total energy is conserved for free target but not for fixed target.
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
For a free target, the target can recoil, sharing kinetic energy with the projectile, leading to less relative kinetic energy converted into potential energy, hence a larger distance of closest approach. Total energy is always conserved if only conservative forces are present for both free and fixed targets. Thus (A) is true and (R) is false.
Assertion (A): When two positive point charges move away from each other, their electrostatic potential energy decreases.
Reason (R): Change in potential energy between two points is equal to the work done by electrostatic forces.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): Electrostatic potential energy between two positive charges is \(U = k \frac{q_1 q_2}{r}\). If they move away, \(r\) increases, thus \(U\) decreases. So (A) is true.
Reason (R): The change in potential energy is defined as the negative of the work done by conservative forces (like electrostatic force): \(Delta U = -W_{electrostatic}\). So (R) is false.nTherefore, (A) is true and (R) is false.
Assertion (A): Electric potential of earth is taken as zero.
Reason (R): Electric field strength on the surface of earth is zero.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): For practical purposes in circuit analysis and grounding, the Earth's electric potential is taken as zero because it is a large conductor and can absorb or supply charge without significant potential change. So (A) is true.
Reason (R): The electric field strength on the Earth's surface is generally not zero; it varies due to atmospheric conditions, presence of charges, etc. So (R) is false.nTherefore, (A) is true and (R) is false.