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Two charges +3.2×10^-19C and -3.2×10^-19C placed 2.4 Å apart form an electric dipole. It is placed in a uniform electric field of intensity 4×10⁵V /m the work done to rotate the electric dipole from the equilibrium position by 180° is

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$3 \times 10^{- 23} J$

$6 \times 10^{- 23} J$

$12 \times 10^{- 23} J$

zero

answer is B.

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Detailed Solution

$W = {PE}_{2} - {PE}_{1} = 2 PE = - 2 \times 2 qdE$

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