[YMP/EM-04008] Force Between the Hemispheres Parts of a Uniformly Charged Sphere

A conducting spherical shell of radius \(R\) is placed in a uniform electric field along the \(Z\)- axis. A plane  perpendicular to the electric field divides the shell into two equal halves. Show that  the force acting on the hemispherical surfaces which tends to separate them is  across the diametral plane perpendicular to \(\vec E\) is given by \[ F = \tfrac{9}{4}\pi \epsilon_0 R^2 E_0^2\] \input{em-que-04046} \subsubsection*{Solution}\begin{small} First we should find the electric field outside the shell, of radius \(R\), and the induced charge density on the surface. We take the known result for the electric field due to the shell in a uniform electric field. From Griffiths, we have the answer for  the potential outside the shell equal to  \begin{equation} \phi(r, \theta)= -E_0\Big(r- \frac{R^3}{r^2}\cos\theta\Big). \end{equation} We taking the answer for the electric field on the surface of the shell as above \(\vec E =(0,0, 3 E_0\cos\theta)\) on the surface of the shell. The force on the hemisphere can be computed  by integrating the Maxwell stress tensor over the surface of the hemisphere. The Maxwell stress tensor is given  by \begin{equation} T_{ij}= \epsilon_0\Big(E_iE_j -\frac{1}{2} \delta_{ij} E^2\Big) \end{equation} in 1-2-3 notation. The normal to the surface of the hemisphere is \[\hat n=\hat{r}= (\sin\theta\cos \phi, \sin\theta \sin\phi, \cos\theta)\] The force on a small  surface element of the hemisphere is given by \begin{equation} \Delta F_j = \sum_{k=1}^3 T_{jk} \hat n_k (\Delta S) \end{equation} Only the \(z\) component of the force is nonzero. Writing the \(z\)- component explicitly, we can see that \(T_{zx}T_{zy}\) vanish \begin{equation} T_{zx}=0, \quad T_{zy}=0,\quad T_{zz} = \frac{9\epsilon_0}{2}E_)^2 \cos^2\theta. \end{equation} Thus the force on a surface element with \(\theta, \phi\)in the ranges \(\theta, \theta+d\theta)\) and \(\phi, \phi+d\phi\) respectively, is given by

\begin{eqnarray}
 \Delta F_z &=& (T_{zx} n_x + T_{zy} n_y + T_{zz} n_z)\\
 &=& T_{zz} n_z = \frac{9}{2}\epsilon_0 E_0^2 \cos^3\theta \sin\theta d\theta d\phi.
\end{eqnarray}
Integrating over \(\theta\) form 0 to \(\pi/2\) and over \(\phi\) from 0 to 2 \(\pi\), we  get
\begin{equation}
 F_z = \int_0^{2\pi}d\phi \int_0^{\pi}
 \frac{9}{2}\epsilon_0 E_0^2 \cos^3\theta \sin\theta\, d\theta= \frac{9}{4}\pi \epsilon_0 R^2 E_0^2.
\end{equation}