Question 638: Electrostatics & Capacitance

Question

An electric dipole of dipole moment is $$6.0 	imes 10^{-6} ~\mathrm{C m}$$ placed in a uniform electric field of $$1.5 	imes 10^{3} ~\mathrm{NC}^{-1}$$ in such a way that dipole moment is along electric field. The work done in rotating dipole by $$180^{\circ}$$ in this field will be ___________ $$\mathrm{m J}$$.

Accepted Answer

The work done $W$ in rotating an electric dipole in a uniform electric field is given by:
$W = pE(1 - \cos	heta)$,
where $p$ is the dipole moment, $E$ is the strength of the electric field, and $	heta$ is the angle the dipole is rotated through.
In this case, the dipole moment $p$ is $6.0 	imes 10^{-6} ~\mathrm{C m}$, the electric field $E$ is $1.5 	imes 10^{3} ~\mathrm{NC}^{-1}$, and the angle $	heta$ is $180^{\circ}$.
Substituting these values into the formula gives:
$W = 6.0 	imes 10^{-6} ~\mathrm{C m} 	imes 1.5 	imes 10^{3} ~\mathrm{NC}^{-1} 	imes (1 - \cos180^{\circ})$.
Since $\cos180^{\circ} = -1$, the equation becomes:
$W = 6.0 	imes 10^{-6} ~\mathrm{C m} 	imes 1.5 	imes 10^{3} ~\mathrm{NC}^{-1} 	imes (1 - (-1))$,
$W = 6.0 	imes 10^{-6} ~\mathrm{C m} 	imes 1.5 	imes 10^{3} ~\mathrm{NC}^{-1} 	imes 2$,
$W = 18.0 	imes 10^{-3} ~\mathrm{J} = 18.0 ~\mathrm{mJ}$.
So the work done in rotating the dipole by $180^{\circ}$ in this field is 18.0 millijoules.

Electrostatics & Capacitance

Integer / Numeric Answer Type