- JEE Mains 2019
- Hard
- Numerical
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......\(V\)
Answer & Solution
(D) \(1\)
Step-by-step Solution
Detailed explanation
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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