A) V
B) 2V
C) -2V
D) 4V
Correct Answer: A
Solution :
[a] As given in the first condition : |
Both conducting spheres are shown. |
\[{{V}_{in}}-{{V}_{out}}=\left( \frac{kQ}{{{r}_{1}}} \right)-\left( \frac{kQ}{{{r}_{2}}} \right)\] |
\[=kQ\left( \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right)=V\] |
In the second condition : |
Shell is now given charge -4Q. |
\[{{V}_{in}}-{{V}_{out}}=\left( \frac{kQ}{{{r}_{1}}}-\frac{4kQ}{{{r}_{2}}} \right)-\left( \frac{kQ}{{{r}_{2}}}-\frac{4kQ}{{{r}_{2}}} \right)\] |
\[=\frac{kQ}{{{r}_{1}}}-\frac{kQ}{{{r}_{2}}}\] |
\[=kQ\left( \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right)=V\] |
Hence, we also obtain that potential difference does not depend on charge of outer sphere. |
\[\therefore \] P.d. remains same |
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