As shown in the figure,a charge +q is moved f | vedclass

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(C) The electric potential due to a dipole at a point $(r, 	heta)$ is given by $V = \frac{1}{4\pi \varepsilon_0} \frac{p \cos 	heta}{r^2}$.Potential

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$\frac{1}{4\pi \varepsilon_0} \frac{\sqrt{2} qp}{r^2}$

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(C) The electric potential due to a dipole at a point $(r, 	heta)$ is given by $V = \frac{1}{4\pi \varepsilon_0} \frac{p \cos 	heta}{r^2}$.Potential at point $A$ $(r, 135^{\circ})$:$V_A = \frac{1}{4\pi \varepsilon_0} \frac{p \cos 135^{\circ}}{r^2} = \frac{1}{4\pi \varepsilon_0} \frac{p (-1/\sqrt{2})}{r^2}$.Potential at point $B$ $(r, 45^{\circ})$:$V_B = \frac{1}{4\pi \varepsilon_0} \frac{p \cos 45^{\circ}}{r^2} = \frac{1}{4\pi \varepsilon_0} \frac{p (1/\sqrt{2})}{r^2}$.The work done by an external agent in moving a charge $q$ from $A$ to $B$ is $W = q(V_B - V_A)$.$W = q \left[ \frac{1}{4\pi \varepsilon_0} \frac{p}{r^2} \left( \frac{1}{\sqrt{2}} - \left( -\frac{1}{\sqrt{2}} ight) ight) ight]$$W = q \left[ \frac{1}{4\pi \varepsilon_0} \frac{p}{r^2} \left( \frac{2}{\sqrt{2}} ight) ight]$$W = \frac{1}{4\pi \varepsilon_0} \frac{\sqrt{2} qp}{r^2}$.

As shown in the figure,a charge $+q$ is moved from point $A \,(r, 135^{\circ})$ to point $B \,(r, 45^{\circ})$. If the dipole moment is $p$,what is the work done by an external agent?

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