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
B
5.8eV
C
13J
D
13eV

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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