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): 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.
A point charge is placed at origin. Assuming potential to be zero at infinity, potential difference at two point A and B is found to be 10 V i.e., \(V_A – V_B = 10 \text{V}\). Now if the reference at infinity is changed to 10 V, then \(V_A – V_B\) will be
View Answer
The potential difference between two points is independent of the reference level of potential because the change in reference shifts both potentials by the same constant value.
If electric potential (in volt) in a region is expressed as \(V(x, y, z) = 2xy – yz\), the electric field (in N/C) at point \((1, 0, 1)\text{ m}\) will be equal to
1. \(-\hat{j}\)
2. \(2\hat{i} - \hat{j}\)
3. \(-\hat{j} + \hat{k}\)
4. \(\hat{i} - \hat{j} + \hat{k}\)
View Answer
Using \(\vec{E} = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)\), we get \(E_x = -2y\), \(E_y = z - 2x\), and \(E_z = y\). Substituting \((1, 0, 1)\), we get \(vec{E} = -\hat{j}\).