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Question

Two point charges +3μC and +8μC repel each other with a force of 40N. If a charge of 5μC is added to each of them, the force between them will become:

A
10N
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B
10N
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C
20N
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D
20N
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Solution

The correct option is A 10N
Redistribution of charges takes place.
Charge q1=3μC
Charge q2=8μC
When third charge q3=5μC is added to each, then new charges on q1 and q2 will be
q1=35=2μC
and, q2=85=3μC
Now,
In first case,
40=14πε03×8r2
In second case,
F=14πε0×(2×3)r2
F40=2×33×8
or F=10N

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