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Imagine that I have a particle of charge $q$ at the center of a spherical insulating shell of charge $Q$ and radius $R$.

Both the particle and shell are initially at rest.

Now I apply a force $\mathbf{F}$ to the particle which causes it to have an acceleration $d\mathbf{v}/dt$.

The electric field due to the charge $q$ is given by:

$$\mathbf{E} = - \mathbf{\nabla} \phi - \frac{\partial \mathbf{A}}{\partial t}.$$

Due to the symmetry of the situation the total static electric force on the shell due to the charge is zero.

The $\mathbf{A}$-field due to the charge $q$ at the position of the shell is approximately given by:

$$\mathbf{A} = \frac{q\mathbf{v}}{4 \pi \epsilon_0 c^2R}.$$

Therefore the charge $q$ induces an electric force $\mathbf{f_s}$ on the shell given by:

$$\mathbf{f_s} = -\frac{qQ}{4 \pi \epsilon_0 c^2R}\frac{d\mathbf{v}}{dt}.$$

But the total force on the system as a whole should remain $\mathbf{F}$.

Does this mean that the shell must induce a balancing electric force $\mathbf{f_s}$ back on the accelerating particle?

P.S. As the particle acceleration is constant I believe that there is no radiation reaction force back on the particle (a controversial view I admit, see related link below). Thus there is no change of momentum in the EM field. Therefore the internal forces on the system of charges should balance.

John Eastmond
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  • related: http://physics.stackexchange.com/q/70915/ –  Sep 07 '13 at 22:34
  • This induces an electric force fs on the shell. Why? –  Sep 07 '13 at 22:35
  • The PS is not necessarily accurate; see the link above. In the revised version of the question, you refer to the vector potential, but the equation you give for it is wrong, and you're also treating it as a scalar when it's actually a vector. Is the shell a conductor? An insulator? –  Sep 07 '13 at 22:58
  • OK, now your equation for A makes more sense, but you're still treating it as a scalar when it's a vector. You might want to delete the question, work it out more carefully, and then post it again when you have all the mistakes straightened out. –  Sep 07 '13 at 23:02
  • I think I've got it right now! – John Eastmond Sep 07 '13 at 23:27
  • http://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential –  Sep 07 '13 at 23:50
  • I could have used the Lienard-Wiechert potential for a more accurate expression for the electric field due to an accelerating charge. It only introduces a geometrical factor of 2/3 to the force on the shell. – John Eastmond Sep 08 '13 at 00:18

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