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Two positive charges 12 $μ$ C and 8 $μ$ C are 10 cm apart. The work done is bringing them 4 cm closer is

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5.8 J

5.8 eV

13 J

13 eV

answer is A.

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Detailed Solution

$\begin{array}{l} W = {PE}_{f} - {PE}_{i} = {kq}_{1} q_{2} (\frac{1}{r_{2}} - \frac{1}{r_{1}}) \\ ∴ W = 9 \times 10^{9} (12 \times 8 \times 10^{- 12}) (\frac{1}{0.06} - \frac{1}{0.10}) \\ ∴ W = 5.8 J \end{array}$

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