A) Increase continuously
B) Decrease continuously
C) First increase, then decrease
D) First decrease, then increase
Correct Answer: C
Solution :
| Let charge on smaller sphere be x and on larger sphere be \[4q-x\] |
| Force between them is given by |
| \[F=\frac{kx\,\,(4q-x)}{{{d}^{2}}}\] |
| \[\frac{dF}{dx}=0\Rightarrow 4q-2x=0\Rightarrow x=2q\] |
 |
| \[\frac{{{d}^{2}}F}{d{{x}^{2}}}=\frac{K}{{{d}^{2}}}(-\,2)<0\] |
| \[\therefore \]It represents a maximum. |
| Final charges on the smaller sphere and the larger sphere are q & 3q respectively as required by the equality of potentials |
| \[\therefore \]Force will increase until the charges becomes equal and after that force will decreases. |
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