[NOTES/EM-04015] Force on a Charged Conductor in Uniform External Electric field

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 \[ F = \frac{9}{4}\pi \epsilon_0 R^2 E_0^2\]

 The charge density on the surface is given by \begin{eqnarray} \sigma &=& -\epsilon_0\frac{\partial \phi}{\partial n} =-\epsilon_0\frac{\partial \phi}{\partial r}\big|_{r=R}\\ &=& 3\epsilon_0 E_0 \cos \theta. \end{eqnarray}

 The electric field just outside the surface is normal to the surface and is given by \begin{equation} \vec E = \frac{\sigma }{\epsilon_0} \hat{r}. \end{equation}

 f we consider a small surface area \(\Delta S\) of the shell, the charge on the surface is \(Delta Q=\sigma \Delta S\) and the force

 \begin{equation} \frac{1}{2} \vec E (\Delta Q)=\frac{1}{2}\vec E (\sigma \Delta S)= \frac{\sigma^2}{2\epsilon_0} \hat{n}\qquad
\end{equation}Why 1/2?}.

where \(\hat n\) is normal to the surface. Thus the force on surface element \(d\Delta S\) will be will be radially outwards and

 \begin{equation}\label{EQ01} d\vec F = \frac{\sigma^2 dS} {2\epsilon_0} \hat r. \end{equation}

when integrated over the hemisphere, the components of the force other than the \(Z\) component will vanish.

Taking the \(Z\)- component of force in \eqref{EQ01} and integrating we get

\begin{equation} dF_z= \frac {\sigma^2\cos\theta }{2\epsilon_0}(2\pi \sin\theta d\theta) \end{equation}

Integrating over \(\theta\) from 0 to \(\pi\) gives the required force on one of the hemispheres as

\begin{eqnarray} F &=&\int_0^{\pi/2} \frac {(\sigma^2)\cos\theta }{2\epsilon_0}(2\pi R^2 \sin\theta d\theta)\\ &=& 9\pi \epsilon_0R^2 \int_0^{\pi/2} \cos^3\theta \sin\theta \, d\theta.\\ &=&\tfrac{9}{4}\pi\epsilon_0 R^2 E_0^2. \end{eqnarray}