Question

An insulating sphere of radius R is coated with a uniform surface charge density $\sigma$ on its exterior surface.

Case-I :-

Suppose that we cut off a spherical cap corresponding to a half-angle $\alpha$ from a certain axis (i.e. the cap has base radius R sin$\alpha$) and remove the rest of the sphere.

Case-II :-

A spherical wedge of half-angle $\alpha$ (a slice of watermelon) is extracted from the sphere and the rest of the sphere is removed instead.

Answer 1

Case I: $Q_{\rm cap}=2\pi\sigma R^2(1-\cos\alpha)$ and
     ${\bf E}_{\rm cap}=\dfrac{\sigma\sin^2\alpha}{4\epsilon_0}\,\hat z.$
 Case II: $Q_{\rm wedge}=4\alpha\,\sigma R^2$ and
     ${\bf E}_{\rm wedge}=-\dfrac{\sigma\sin\alpha}{4\epsilon_0}\,\hat x.$

Answer 2

• For the cap, use the area formula $A_{\rm cap}=2\pi R^2(1-\cos\alpha)$ to get the net charge. Then find the field at the centre by integrating the contribution

$dE_z=\frac{\sigma}{4\pi\epsilon_0}\sin\theta\cos\theta\,d\theta\,d\varphi$,

giving

$E_z=\frac{\sigma\sin^2\alpha}{4\epsilon_0}.$

• For the wedge, note that its area is the fraction $\frac{2\alpha}{2\pi}$ of the total area so that $A_{\rm wedge}=4\alpha R^2$. Then, by symmetry, only the $x$–components add (with the $z$–components cancelling) and one obtains

$E_x=-\frac{\sigma\sin\alpha}{4\epsilon_0}.$