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?


  • $\begingroup$ Duplicate of physics.stackexchange.com/q/228720 $\endgroup$
    – velut luna
    Apr 11, 2016 at 8:39
  • 1
    $\begingroup$ $$\begin{array}{cc} \ & \begin{array}{cc} \frac{Q}{4 \pi r^2 \epsilon _0} & r>=R \\ 0 & r<R \\ \end{array} \\ \end{array}$$ $\endgroup$ Apr 22, 2016 at 13:34

1 Answer 1


Those formulas, and those graphs are idealizations. In reality, there are no discontinuous fields, just as there are no zero-thickness shells. If you start with an impossible situation, you will calculate impossible and meaningless results. For real situations, situations for which the theory is valid, the field may change quickly, but it does so smoothly. Your question, "what would the field be at that distance" has no answer. Let me ask you this: how fast can a unicorn run?


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