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Two charged particles A and B are kept at a certain separation and magnitude of electric force acting on each particle is F. Now additional 50% of charge of A is added to A and the separation between A and B is halved. Now what will be the magnitude of electric force acting on each particle?

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4 F

$\frac{8 F}{3}$

$\frac{4 F}{3}$

6 F

answer is D.

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Detailed Solution

$F = K \frac{Q_{1} Q_{2}}{r^{2}}$ Here $Q_{1} =$ charge on A and $Q_{2} =$ charge on B

$Q_{1}^{1} = Q_{1} + 50 % of Q_{1} = \frac{3}{2} Q_{1}$

$∴ F^{1} = K \frac{Q_{1}^{1} . Q_{2}}{{(\frac{r}{2})}^{2}} = K \frac{\frac{3 Q_{1}}{2} . Q_{2}}{\frac{r^{2}}{4}} = 6 \frac{K Q_{1} Q_{2}}{r^{2}}$

$\Rightarrow F^{1} = 6 F$

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