Two electric dipoles,each of dipole moment P, | vedclass

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(B) The potential energy of an electric dipole in an external electric field $E$ is given by $U = -\vec{p} \cdot \vec{E} = -pE \cos 	heta$,where $	h

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(B) The potential energy of an electric dipole in an external electric field $E$ is given by $U = -\vec{p} \cdot \vec{E} = -pE \cos 	heta$,where $	heta$ is the angle between the dipole moment $\vec{p}$ and the electric field $\vec{E}$.From the figure,the dipole at $A$ is oriented along the $+x$-direction $(	heta_A = 0^{\circ})$ and the dipole at $B$ is oriented along the $+y$-direction $(	heta_B = 90^{\circ})$.Initial potential energy: $U_i = (-pE \cos 0^{\circ}) + (-pE \cos 90^{\circ}) = -pE + 0 = -pE$.After rotating both dipoles by $90^{\circ}$ clockwise:The dipole at $A$ (initially at $0^{\circ}$) rotates to $-90^{\circ}$ (or $270^{\circ}$).The dipole at $B$ (initially at $90^{\circ}$) rotates to $0^{\circ}$.Final potential energy: $U_f = (-pE \cos(-90^{\circ})) + (-pE \cos 0^{\circ}) = 0 - pE = -pE$.The work done $W$ is the change in potential energy: $W = U_f - U_i = (-pE) - (-pE) = 0$.

Two electric dipoles,each of dipole moment $P$,are placed at points $A(a, 0)$ and $B(-a, 0)$ as shown in the figure. The work done in rotating both the dipoles through $90^{\circ}$ in the clockwise direction is $(E = \text{Electric field})$

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