Question

A solid conducting sphere having a charge $Q$ is surrounded by an uncharged concentric conducting spherical shell. The potential difference between the surface of solid sphere and the shell is $V$. The shell is now given a charge $-3Q$. The new potential difference between the same surfaces will be :

• A. $-2V$
• B. $4V$
• C. $V$
• D. $2V$

Answer 1

The correct option is C $V$

A solid conducting sphere having a charge Q is surrounded by an uncharged concentric spherical shell, The potential difference between the solid sphere and hallow shell is $V$.

$V_A-V_B=V$

$V_A-V_B=\frac{Q}{4\pi {\epsilon }_{0}a}.\frac{Q}{4\pi {\epsilon }_0.b}$

$V=\frac{Q}{4\pi {\epsilon }_0}[\frac{1}{a}-\frac{1}{b}]$. . . . . . .(1)

Now, shell is given a charge $-3Q$,the,

The new potential is,

$V_A^{\prime }=\frac{Q}{4\pi {\epsilon }_0.a}-\frac{3Q}{4\pi {\epsilon }_0.b}$

$V_A^{\prime }=\frac{Q}{4\pi {\epsilon }_0}[\frac{1}{a}-\frac{3}{b}]$ and,

$V_B^{\prime }=\frac{Q}{4\pi {\epsilon }_{0}b}-\frac{3Q}{4\pi {\epsilon }_{0}b}$

$V_B^{\prime }=\frac{Q}{4\pi {\epsilon }_0}[\frac{1}{b}-\frac{3}{b}]$

$V_A^{\prime }-V_B^{\prime }=\frac{Q}{4\pi {\epsilon }_0}[\frac{1}{a}-\frac{1}{b}]=V$

The correct option is C.