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(C) The electric potential $V$ due to a dipole at a point $(r, 	heta)$ is given by $V = \frac{kp \cos 	heta}{r^2}$.For the equatorial plane (the $y$

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The work done in bringing a test charge from point $A$ to $C$ is non-zero

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(C) The electric potential $V$ due to a dipole at a point $(r, 	heta)$ is given by $V = \frac{kp \cos 	heta}{r^2}$.For the equatorial plane (the $y$-axis in this figure),$	heta = 90^\circ$,so $\cos 90^\circ = 0$,which means the potential is zero at all points on the $y$-axis,including points $B$ and $D$.Since $V_B = 0$ and $V_D = 0$,the work done in moving a test charge $q_0$ from $B$ to $D$ is $W = q_0(V_D - V_B) = 0$.For points $A$ and $C$ on the axial line,the potential is non-zero and $V_A 
eq V_C$. Therefore,the work done in moving a charge between $A$ and $C$ is non-zero.

An electric dipole is placed on the $x$-axis as shown in the figure. Consider a spherical closed surface around it. Then,

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