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Question

Two charges of equal magnitude q are placed in air at a distance 2a apart and third charge 2q is placed at mid-point. The potential energy of the system is (ε0= permittivity of free space) :

A
q28πε0a
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B
3q28πε0a
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C
5q28πε0a
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D
7q28πε0a
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Solution

The correct option is D 7q28πε0a
Potential energy of the system
U=14πε0[q1q2r12+q1q3r13+q2q3r23]
=14πε0[q(2q)a+q(2q)a+qq2a]
=14πε0[2q2a2q2a+q22a]
=14πε0[4q2a+q22a]=7q28πε0a

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