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Q.

Two identical metal balls with charges + 2 Q and -Q are separated by some distance and exert a force F on each other. They are joined by a conducting wire, which is than removed. The force between them will now be

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a

F/4

b

F/2

c

F

d

F/8

answer is D.

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Detailed Solution

$F = \frac{1}{4 {πε}_{0}} \times \frac{2 Q \times Q}{r^{2}} = \frac{1}{4 {πε}_{0}} \times \frac{2 Q^{2}}{r^{2}}$
After joining by a conducting wire, the charge on each metal ball will be
$(2 Q - Q) / 2 = Q / 2$
$\begin{array}{l} ∴ F^{'} = \frac{1}{4 {πε}_{0}} \times \frac{(Q / 2) (Q / 2)}{r^{2}} \\ = \frac{1}{4 {πε}_{0}} (\frac{Q^{2}}{4 r^{2}}) = \frac{F}{8} \end{array}$

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