A solid conducting sphere, having a charge Q, | vedclass

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Question

As given in the first condition:

Both conducting spheres are shown.

${V_{in}} - {V_{{	ext{out }}}} = \left( {\frac{{{	ext{kQ}}}}{{{r_1}}}} i

Vedclass · Accepted Answer

$1$

Vedclass · Answer

As given in the first condition:

Both conducting spheres are shown.

${V_{in}} - {V_{{	ext{out }}}} = \left( {\frac{{{	ext{kQ}}}}{{{r_1}}}} ight) - \left( {\frac{{{	ext{kQ}}}}{{{{	ext{r}}_2}}}} ight)$

$ = {	ext{kQ}}\left( {\frac{1}{{{{	ext{r}}_1}}} - \frac{1}{{{{	ext{r}}_2}}}} ight) = V$

In the second condition:

Shell is now given charge $-4 Q$ 

${V_{in}} - {V_{out}} = \left( {\frac{{kQ}}{{{r_1}}} - \frac{{4kQ}}{{{r_2}}}} ight) - \left( {\frac{{kQ}}{{{r_2}}} - \frac{{4kQ}}{{{r_2}}}} ight)$

$ = \frac{{kQ}}{{{r_1}}} - \frac{{kQ}}{{{r_2}}}$

$=\mathrm{kQ}\left(\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}ight)=\mathrm{V}$

Hence, we also obtain that potential difference does not depend on charge of outer sphere.

$	herefore $ $P.d.$ remains same.

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