Two point charges +3 μC and +8 μC repel each other with a force of 40 N. If a charge of -5 μC is added to each of them, then the force between them will become

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-10 N

+10 N

+20 N

-20 N

answer is A.

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Detailed Solution

In second case, charges will be -2 μC and + 3 μC

Since F $\propto Q_{1} Q_{2}$ , we have

$\begin{array}{l} \frac{F}{F'} = \frac{Q_{1} Q_{2}}{Q_{1}^{'} Q_{2}^{'}} \\ ∴ \frac{40}{F'} = \frac{3 \times 8}{- 2 \times 3} = - 4 \Rightarrow F' = - 10 N (Attractive) \end{array}$

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