A solid conducting sphere having a charge Q i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the radius of the solid sphere be $r_a$ and the radius of the hollow shell be $r_b$.The potential at the surface of the solid sphere is $V_{sp

Vedclass · Accepted Answer

$V$

Vedclass · Answer

(A) Let the radius of the solid sphere be $r_a$ and the radius of the hollow shell be $r_b$.The potential at the surface of the solid sphere is $V_{sphere} = \frac{KQ}{r_a} + \frac{K(0)}{r_b} = \frac{KQ}{r_a}$.The potential at the surface of the hollow shell is $V_{shell} = \frac{KQ}{r_b} + \frac{K(0)}{r_b} = \frac{KQ}{r_b}$.The initial potential difference is $V = V_{sphere} - V_{shell} = K Q \left( \frac{1}{r_a} - \frac{1}{r_b} \right)$.When the shell is given a charge $-3Q$,the new potential at the surface of the solid sphere is $V'_{sphere} = \frac{KQ}{r_a} + \frac{K(-3Q)}{r_b}$.The new potential at the surface of the hollow shell is $V'_{shell} = \frac{KQ}{r_b} + \frac{K(-3Q)}{r_b} = -\frac{2KQ}{r_b}$.The new potential difference is $V' = V'_{sphere} - V'_{shell} = \left( \frac{KQ}{r_a} - \frac{3KQ}{r_b} \right) - \left( -\frac{2KQ}{r_b} \right) = \frac{KQ}{r_a} - \frac{KQ}{r_b} = V$.Thus,the potential difference remains unchanged.

Similar Questions

$A$ charge $(-q)$ and another charge $(+Q)$ are kept at two points $A$ and $B$ respectively. Keeping the charge $(+Q)$ fixed at $B$,the charge $(-q)$ at $A$ is moved to another point $C$ such that $ABC$ forms an equilateral triangle of side $l$. The net work done in moving the charge $(-q)$ is

$A$ thin spherical shell is charged by some source. The potential difference between the two points $C$ (the center) and $P$ (a point on the surface) as shown in the figure is: (Take $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ SI units}$)

The radius of a soap bubble whose potential is $16\,V$ is doubled. The new potential of the bubble will be.....$V$

The figure shows the electric field lines of a positive and a negative charge,respectively. Determine the sign of the potential difference $V_Q - V_P$ and $V_B - V_A$.

When a charge of $20 C$ is taken from one point to another separated by a distance of $0.2 m$,work of $2 J$ is required to be done. What is the potential difference between the two points?

Vedclass Test Series

Exam Paper Generator

Online Exam Module