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Two charges +3.2×1019 C and 3.2×1019 C separated by 2r=2.4 ˚A forms an electric dipole. It is placed in a uniform electric field of 4×105 V/m. The work done by electric force to rotate the electric dipole from the equilibrium position to 180 is

A
6×1023 J
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B
6×1023 J
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C
3×1023 J
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D
Zero
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Solution

The correct option is A 6×1023 J
Given:

2r=2.4 ˚A=2.4×1010 m <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->

E=4×105 V/m

q=3.2×1019 C

We know, p=q(2r)

Also, work done by electric force,

W=pE(cosθ2cosθ1) ...(1)

θ1=0; cos0=1θ2=180; cos180=1

From (1), we get,

W=q(2r)E(cos180cos0)

=3.2×1019×2.4×1010×4×105×(11)

W6×1023 J

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