An electric dipole consists of two charges of 3.210^-19 C and -3.210^-19C separated by 2.410^-10 m.the dipole is placed in a uniform electric field of 410^5 V/m.Calculate the work done in rotating the dipole through 180 from the equilibrium position.

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Solution

Dear student,

The work done in rotating the dipole = change in potential energy of dipole.
Potential energy, $U = - p E \cos θ$
where $θ$ is the angle between electric field and dipole moment.
At equilibrium $θ = 0 °$
$U_{i n i t i a l} = - p E$
$U_{f i n a l} = - p E \cos (180) = p E$
So work done
$U_{f i n a l} - U_{i n i t i a l} = 2 p E = 2 |q| l E = 2 \times (3.2 \times 10^{- 19}) \times (2.4 \times 10^{- 10}) \times (4 \times 10^{5}) = 6.144 \times 10^{- 23} J$
Regards

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