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Two positive point charges of $12 μC$ and $8 μC$ are 10cm apart. The work done in bringing them 4 cm closer is

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5.8 eV

5.8 J

13 J

13 eV

answer is C.

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Detailed Solution

$W = U_{f} - U_{i} = 9 \times 10^{9} \times Q_{1} Q_{2} [\frac{1}{r_{2}} - \frac{1}{r_{1}}]$

$\Rightarrow W = 9 \times 10^{9} \times 12 \times 10^{- 6} \times 8 \times 10^{- 6} [\frac{1}{4 \times 10^{- 2}} - \frac{1}{10 \times 10^{- 2}}]$

$= 12 .96 J \approx 13 J$

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