Potential due to a dipole placed at the centre of a sphere Recently I came across a question which stated that a small electric dipole is placed at the centre of a sphere(insulated). I'm required to prove that the potential due to this dipole is always zero on a circle on the surface of the sphere. What does a circle on the surface of a sphere mean here? Is it the planar section cut off by a plane and a sphere? I know that for an equatorial point with respect to a dipole, potential is always zero. But how can a circle be always at an equatorial position with respect to the dipole at the centre of sphere? Can someone help me in this and illustrate it using a diagram?
 A: I think you may already have the correct intuition here. If the dipole is oriented towards the north pole of the sphere, the circle around the equator of the sphere (i.e. if you draw a line all the way around the sphere at the equator) has a potential of zero.
A intuitive argument for why the potential is zero is easy: in the plane perpendicular to the dipole moment, the electric field is always antiparallel to the dipole moment, so if you were moving a charge $q$ in from $r=\infty$ toward the dipole, the work done by the electric field would be $\vec E \cdot \vec{dl}=0$ because $\vec E$ is always perpendicular to $\vec{dl}$.
The field of a dipole $\vec p$ at position $\vec R$:
$$ \vec E (\vec R) = \frac{3 (\vec p \cdot \hat R) \hat R - \vec p }{4\pi \epsilon_0 R^3}$$
If we set $\vec p = p_0 \hat z $ and $\vec R = r \hat x$, then we get
$$ \vec E = \frac{3 p_0 (\hat z \cdot \hat x) \hat x - p_0 \hat z }{4\pi \epsilon_0 R^3} = -\frac{ p_0 \hat z }{4\pi \epsilon_0 R^3} $$
For any point final point $\vec R_f = r_f \hat x$, with $\vec{dl} = dR \hat x$ the potential is
$$ V = - \int \limits_{\infty}^{r_f} \vec E \cdot \vec{dl} = \int \limits_{\infty}^{r_f} \frac{ p_0 \hat z \cdot dR \hat x }{4\pi \epsilon_0 R^3} = 0 $$
