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Question

(A) The work done by an external electric field on an electric dipole is given by the formula: $W_{	ext{field}} = pE(\cos 	heta_1 - \cos 	heta_2)$.

Vedclass · Accepted Answer

$\frac{pE}{2}(\sqrt{3} - 1)$

Vedclass · Answer

(A) The work done by an external electric field on an electric dipole is given by the formula: $W_{	ext{field}} = pE(\cos 	heta_1 - \cos 	heta_2)$.Alternatively,if we consider the work done by the field as the negative change in potential energy,$W_{	ext{field}} = -(U_2 - U_1) = U_1 - U_2 = pE(\cos 	heta_1 - \cos 	heta_2)$.Given: $	heta_1 = 30^\circ$ and $	heta_2 = 60^\circ$.Substituting the values:$W_{	ext{field}} = pE(\cos 30^\circ - \cos 60^\circ)$$W_{	ext{field}} = pE\left(\frac{\sqrt{3}}{2} - \frac{1}{2}ight)$$W_{	ext{field}} = \frac{pE}{2}(\sqrt{3} - 1)$.

An electric dipole of moment $\vec{p}$ is kept in a uniform electric field $\vec{E}$. The angle between $\vec{p}$ and $\vec{E}$ is $30^\circ$. Calculate the work done by the field when the angle is increased to $60^\circ$.

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