# Electric field at surface of a spherical shell

The shell theorem provides a well known result that for a spherical shell with uniformly distributed charge $Q$ and radius $R$, the electric field at a distance of $r$ from the center is:

$$\begin{array}{cc} \ & \begin{array}{cc} \frac{Q}{4 \pi r^2 \epsilon _0} & r>R \\ 0 & r<R \\ \end{array} \\ \end{array}$$

Or plotted,

However, there appears to be a discontinuity at $r = R$. What would the field be at this distance? In real-life, of course, you cannot lie perfectly on the surface but for a mathematical shell this is of-course valid right?

Also interestingly, the potential (being the integral of the electric field) doesn't suffer from the same discontinuity (though it of course lacks differentiability at $r = R$). Is there any physical significance to this?

• Duplicate of physics.stackexchange.com/q/228720 Apr 11, 2016 at 8:39
• $$\begin{array}{cc} \ & \begin{array}{cc} \frac{Q}{4 \pi r^2 \epsilon _0} & r>=R \\ 0 & r<R \\ \end{array} \\ \end{array}$$ Apr 22, 2016 at 13:34