A) On the surface of the sphere \[B\]
B) On the surface of the sphere \[A\]
C) Equal on both
D) Zero on both
Correct Answer: B
Solution :
After connection of wire, potential becomes equal \ \[\frac{{{Q}_{1}}}{{{r}_{1}}}=\frac{{{Q}_{2}}}{{{r}_{2}}}\] Þ \[\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\] when r1 > r2, then \[{{Q}_{1}}>{{Q}_{2}}\]You need to login to perform this action.
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