Two point charges 4q and 16q are placed at a distance ‘r’ apart. Suppose a third charge –q is placed in between, on the line joining 4q and 16q, so that the electric potential energy of the system is minimum. The position of third charge is

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r/3 from 16q

r/3 from 4q

2r/5 from 16q

2r/3 from 4q

answer is A.

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Detailed Solution

$p o t e n t i a l e n e r g y o f t h e s y s t e m U = - \frac{4 q^{2}}{x} - \frac{16 q^{2}}{r - x} + \frac{64 q^{2}}{r^{2}}$

$i f p o t e n t i a l e n e r g y h a s t o b e m i n i m u m \frac{d u}{d x} = 0 = \frac{4}{x^{2}} - \frac{16}{{(r - x)}^{2}}$

$\begin{array}{rcl} \frac{1}{x} & = & \frac{2}{r - x} \\ x & = & \frac{r}{3} \end{array}$

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