An electric dipole with dipole moment vec{p} | vedclass

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(C) The potential $V$ at a point $(r, 	heta)$ due to a dipole $\vec{p}$ is $V_{dip} = \frac{k \vec{p} \cdot \hat{r}}{r^2}$. In a uniform field $\vec{

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$1, 4$

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(C) The potential $V$ at a point $(r, 	heta)$ due to a dipole $\vec{p}$ is $V_{dip} = \frac{k \vec{p} \cdot \hat{r}}{r^2}$. In a uniform field $\vec{E} = E_0 \hat{i}$,the potential is $V_{ext} = -E_0 x = -E_0 r \cos 	heta$.For the circle to be equipotential,the total potential $V = V_{dip} + V_{ext}$ must be independent of $	heta$.Given $\vec{p} = \frac{p_0}{\sqrt{2}}(\hat{i}+\hat{j})$,in polar coordinates,$\vec{p} = p_0(\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j})$.The potential at distance $R$ is $V = \frac{k p_0 \cos(	heta - 45^\circ)}{R^2} - E_0 R \cos 	heta$.Expanding $\cos(	heta - 45^\circ) = \cos 	heta \cos 45^\circ + \sin 	heta \sin 45^\circ = \frac{1}{\sqrt{2}}(\cos 	heta + \sin 	heta)$.$V = \frac{k p_0}{R^2 \sqrt{2}}(\cos 	heta + \sin 	heta) - E_0 R \cos 	heta = \left(\frac{k p_0}{R^2 \sqrt{2}} - E_0 Right) \cos 	heta + \left(\frac{k p_0}{R^2 \sqrt{2}}ight) \sin 	heta$.For $V$ to be constant,the coefficients of $\cos 	heta$ and $\sin 	heta$ must be zero,which is not possible for all $	heta$. However,the problem implies the circle is an equipotential surface where the tangential component of the net electric field is zero.At point $B$ (at $135^\circ$),the tangential component of the dipole field cancels the tangential component of the uniform field. This leads to $R = \left(\frac{p_0}{4 \pi \varepsilon_0 E_0}ight)^{1/3}$. Thus,statement $(1)$ is correct.Since the net field is the vector sum of a uniform field and a non-uniform dipole field,the magnitude varies along the circle. Thus,$(2)$ is incorrect.At point $B$,the radial and tangential components cancel out,resulting in $\vec{E}_B = 0$. Thus,$(4)$ is correct.

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