The work done to rotate the electric dipole from the equilibrium position by $180^{0}$ is :

3 \times 10^{- 23} J

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6 \times 10^{- 23} J

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12 \times 10^{- 23} J

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Z e r o

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Solution

The correct option is D $6 \times 10^{- 23} J$
We know
work done $= Δ U$ (change in P.E)
$= (- P E c o s θ_{2}) - (- P E c o s θ_{1})$
$= (- P E c o s (180)) - (- P E c o s (0))$
$= P E + P E$
$= 2 P E$
$= 2 \times 7.68 \times 10^{- 27} \times 4 \times 10^{5}$
$W = 6 \times 10^{- 23} J$

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