Question

A metal sphere $\text{A}$ of radius $a$ is charged to potential $V$. What will be the its potential if it is enclosed by a spherical conducting shell $\text{B}$ of radius $b$ and the two are connected by metalic wire.

• A. $\frac{b^{2}V}{a}$
• B. $\frac{a^{2}V}{b}$
• C. $\frac{bV}{a}$
• D. $\frac{a}{b}V$

Answer 1

The correct option is D $\frac{a}{b}V$

The potential of solid sphere $\text{A}$ is,

$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{a}.......\left(1\right)$


After connecting the solid sphere with spherical shell by a metal wire, all the charges will go to the surface of the spherical shell.

The potential of $\text{A}$=The potential of $\text{B}$



$V_A=\frac{1}{4\pi {\epsilon }_0}\frac{q}{b}.......\left(2\right)$

On dividing (2) $÷$ (1)

$\frac{V_A}{V}=\frac{a}{b}$

$V_A=\frac{a}{b}V$

Hence, (d) is the correct answer.