Question

# Two charges +3.2×10−19 C and −3.2×10−19 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 −6×10−23 JGiven: 2r=2.4 ˚A=2.4×10−10 m <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> E=4×105 V/m q=3.2×10−19 C We know, p=q(2r) Also, work done by electric force, W=pE(cosθ2−cosθ1) ...(1) θ1=0∘; cos0∘=1θ2=180∘; cos180∘=−1 From (1), we get, W=q(2r)E(cos180∘−cos0∘) =3.2×10−19×2.4×10−10×4×105×(−1−1) ⇒W≈−6×10−23 J

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