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Two charges 2C and 6C are separated by a finite distance. If a charge of -4C is added to each of them, the initial force of $12 \times 10^{3} N$ will change to

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$6 \times 10^{3} N attraction$

$4 \times 10^{2} N repulsion$

$4 \times 10^{3} N repulsion$

$4 \times 10^{3} N attraction$

answer is D.

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Detailed Solution

$F = \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{r^{2}}$

$\frac{F_{1}}{F_{2}} = \frac{q_{1} q_{2}}{q_{1}' q_{2}'} = \frac{2 x 6}{2 x 2}$

$\Rightarrow F_{2} = 4 x 10^{3} N$

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