Two charged spherical conductors of radius \(R_1\) and \(R_2\) are connected by a wire. Then the ratio of final surface charge densities of the spheres \(\sigma_1 / \sigma_2\) is
1. \(\frac{R_1^2}{R_2^2}\)
2. \(\frac{R_1}{R_2}\)
3. \(\frac{R_2}{R_1}\)
4. \(\sqrt{\frac{R_1}{R_2}}\)
View Answer
When connected, their electric potentials become equal, i.e., \(V_1 = V_2\). Since \(V = \frac{\sigma R}{\varepsilon_0}\), we get \(\sigma_1 R_1 = \sigma_2 R_2\), which gives \(\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}\).
Four electric charges of \(8\ mu\text{C}\), \(5\ mu\text{C}\), \(-3\ mu\text{C}\), \(-10\ mu\text{C}\) are placed at the corners of a square of side \(\sqrt{2}\text{ m}\). The potential at the centre of the square is
1. \(9 \times 10^3 V\)
2. Zero
3. \(1.8 \times 10^3 V\)
4. \(2.7 \times 10^3 V\)
View Answer
The total electric potential at the centre is \(V = \sum \frac{kq_i}{r}\). Since the distance \(r\) from each corner to the centre is equal and the sum of charges \(\sum q_i = 8 + 5 - 3 - 10 = 0\), the net potential is zero.
If some positive charge is given to a solid conductor, then its potential is
1. Minimum at surface
2. Zero at centre
3. Same throughout the conductor
4. Maximum somewhere outside the surface
View Answer
The electric field inside a conductor is zero, so no work is done in moving a charge inside it. Consequently, the potential is constant and same throughout the conductor.
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.
Assertion (A): If a point charge be rotated in a circle around another stationary charge at the centre of the circle, the work done by electric field will be zero.
Reason (R): Work done by centripetal force is always 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
Both (A) and (R) are true. (A) is true because the electric field is conservative and the path is equipotential. (R) is true as centripetal force is perpendicular to displacement (\(W = Fdr\cos 90^\circ = 0\)). However, (R) does not explain (A).
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.
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).