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Question

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


A
5.8J
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B
5.8eV
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C
13J
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D
13eV
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Solution

The correct option is A $$5.8J$$
$$r_1 = 10 cm = 0.1 m$$

Final distance$$ = (10 - 4) cm = 6 cm =0.06 m$$
Work done, W $$= 9 \times 10^9 \times 10^{-6} \times 8 \times 12 \times 10^{-6}[\cfrac{1}{0.06} -\cfrac{1}{0.1}]  = 5.76 J \approx 5.8J$$

Physics

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