A solid conducting sphere, having a charge Q, is surrounded by an… - Vidyadip

MCQ

A solid conducting sphere, having a charge $Q$, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be $V$. If the shell is now given a charge of $-4\, Q$, the new potential difference between the same two surface is......$V$

A
$2$
B
$-2$
C
$4$
✓
$1$

✓

Answer

Correct option: D.

$1$

d
As given in the first condition:

Both conducting spheres are shown.

${V_{in}} - {V_{{\text{out }}}} = \left( {\frac{{{\text{kQ}}}}{{{r_1}}}} \right) - \left( {\frac{{{\text{kQ}}}}{{{{\text{r}}_2}}}} \right)$

$ = {\text{kQ}}\left( {\frac{1}{{{{\text{r}}_1}}} - \frac{1}{{{{\text{r}}_2}}}} \right) = V$

In the second condition:

Shell is now given charge $-4 Q$

${V_{in}} - {V_{out}} = \left( {\frac{{kQ}}{{{r_1}}} - \frac{{4kQ}}{{{r_2}}}} \right) - \left( {\frac{{kQ}}{{{r_2}}} - \frac{{4kQ}}{{{r_2}}}} \right)$

$ = \frac{{kQ}}{{{r_1}}} - \frac{{kQ}}{{{r_2}}}$

$=\mathrm{kQ}\left(\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}\right)=\mathrm{V}$

Hence, we also obtain that potential difference does not depend on charge of outer sphere.

$\therefore $ $P.d.$ remains same.

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