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Question

Two positive point charges are of 12C and 8C are 10cm apart from each other. The work done in bringing them 4cm closer is

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

The correct option is D 13J
Given two charges q1=12μC, q2=8μC are at distance 'd1'=10cm
We have to find the work done in bringing them d2=4cm closer.
For this, we find the charge in the potential energy of the system and then take the difference.
Now d1=10cm=10×102=101m
d2=4cm=4×102m
So, potential (V1) when q1 and q2 are distance d1 apart,
V1=14πε0q1q2d1
Potential (V2) when q1 and q2 are distance d2 apart,
V2=14πε0q1q2d2

Work done =V2V1=14πε0q1q2(1d21d1)
=9×109×12×8×1012(14×102110×102)
=864×103[6×10240×102]
=129.6×101
=12.9613Joules

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