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A ball of mass m = 0.5 kg is suspended by a thread and a charge q = 0.1 µC is supplied. When a ball with diameter 5cm and a like charge of same magnitude is brought close to the first ball, but below it, the tension decreases to 1/3 of its initial value. The distance between centres of the balls is

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$0.12 \times 10^{- 2} m$

$0 .51 \times 10^{- 4} m$

$0.2 \times 10^{- 5} m$

$0.52 \times 10^{- 2} m$

answer is D.

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Detailed Solution

$\begin{array}{l} T + \frac{1}{4 {πε}_{o}} \frac{q_{1} q_{2}}{r^{2}} = mg \end{array}$

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