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Question

Two point charges of 3.2×1019 C and 3.2×1019 C are seperated from each other by 2.4×1010 m. The dipole is situated in a uniform electric field of intensity 4×105 V/m. The work done in rotating the dipole by 180 is p×1024 J. Find the value of p.

A
61.44
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B
61
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C
61.4
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Solution

Given,

q1=3.2×1019 C , q2=3.2×1019 C

Seperation between two charges d=2.4×1010 m

From the data given in the question,

Dipole is placed in a uniform electric field of intensity E=4×105 V/m

Workdone in rotating the dipole by angle θ is given by

W=pE(1cosθ)

But , given θ=180.

W=pE[1(1)]=2pE

W=2(qd)E

Substituting the given values we get,

W=2×3.2×1019×2.4×1010×4×105 J

W=61.44×1024 J

Comparing this with , x×1024 J we get, x=61.44

Accepted answers: 61 , 61.4 , 61.44

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