JEE Main 2023 — Electrostatics Question with Solution

The work done $W$ in rotating an electric dipole in a uniform electric field is given by:

$W=pE(1-\cos \theta )$,

where $p$ is the dipole moment, $E$ is the strength of the electric field, and $\theta$ is the angle the dipole is rotated through.

In this case, the dipole moment $p$ is $6.0\times 10^{-6}\mathrm{Cm}$, the electric field $E$ is $1.5\times 10^3{\mathrm{NC}}^{-1}$, and the angle $\theta$ is $180^{\circ }$.

Substituting these values into the formula gives:

$W=6.0\times 10^{-6}\mathrm{Cm}\times 1.5\times 10^3{\mathrm{NC}}^{-1}\times (1-\cos 180^{\circ })$.

Since $\cos 180^{\circ }=-1$, the equation becomes:

$W=6.0\times 10^{-6}\mathrm{Cm}\times 1.5\times 10^3{\mathrm{NC}}^{-1}\times (1-(-1))$,

$W=6.0\times 10^{-6}\mathrm{Cm}\times 1.5\times 10^3{\mathrm{NC}}^{-1}\times 2$,

$W=18.0\times 10^{-3}\mathrm{J}=18.0\mathrm{mJ}$.

So the work done in rotating the dipole by $180^{\circ }$ in this field is 18.0 millijoules.