# Accelerated charge inside sphere (again!)

Sorry to go on about this scenario again but I think something is going on here.

Imagine a stationary charge $q$, with mass $m$, at the center of a stationary hollow spherical dielectric shell with radius $R$, mass $M$ and total charge $-Q$.

I apply a force $\mathbf{F}$ to charge $q$ so that it accelerates:

$$\mathbf{F} = m \mathbf{a}$$

The accelerating charge $q$ produces a retarded (forwards in time) radiation electric field at the sphere. When integrated over the sphere this field leads to a total force $\mathbf{f}$ on the sphere given by:

$$\mathbf{f} = \frac{2}{3} \frac{qQ}{4\pi\epsilon_0c^2R}\mathbf{a}.$$

So I apply an external force $\mathbf{F}$ to the system (charge + sphere) but a total force $\mathbf{F}+\mathbf{f}$ operates on the system.

Isn't there an inconsistency here?

As the acceleration of charge $q$ is constant there is no radiation reaction force reacting back on it from its electromagnetic field - so that's not the answer.

Instead maybe there is a reaction force back from the charged shell, $-\mathbf{f}$, to the charge $q$ so that the equation of motion for the charge is given by:

$$\mathbf{F} - \mathbf{f} = m\mathbf{a}\ \ \ \ \ \ \ \ \ \ \ (1)$$

This reaction force might be mediated by an advanced electromagnetic interaction going backwards in time from the shell to the charge so that it acts at the moment the charge is accelerated.

Now the total force acting on the system is the same as the force supplied:

$$\mathbf{F} - \mathbf{f} + \mathbf{f} = \mathbf{F}.$$

If one rearranges Equation (1) one gets:

$$\mathbf{F} = (m + \frac{2}{3} \frac{qQ}{4\pi\epsilon_0c^2R}) \mathbf{a}$$

Thus the effective mass of the charge $q$ has increased.

## 1 Answer

There is no obvious inconsistency, whether we use retarded, advanced, or any other field.

If we use only retarded fields, things go as follows. At the time $t=0$, we begin to exert force $\mathbf{F}$ on the charge $q$. It will move with acceleration $\mathbf{F}/m$ for the time interval $R/c$, where $R$ is the radius of the sphere. At the time $t = R/c$, the sphere will be acted upon by the field emanated from the accelerated charge at the time $t=0$.

The fact that the system charge+sphere may experience force greater than the force exerted by us on the charge $q$ is alright. This can happen in EM theory. The total momentum is still conserved, because when sphere gets certain amount of momentum in one direction through its surface charges, the same but oppositely oriented momentum is acquired by the total field at the place where the interaction occurs (the vicinity of these sphere charges).

After the wavefront of the acceleration field of the center charge reaches and moves away from the sphere, the region where the negative momentum is contained will move away with it too.

• Ok. So you're saying that the change in momentum of the sphere is balanced by a change in momentum of the EM field outside the sphere? Maybe that's right. – John Eastmond Oct 14 '13 at 6:49
• If the charge reaches velocity $\Delta v$ then there will be an opposite momentum in the field outside of the sphere given by $-(2/3)(qQ/4\pi\epsilon_0 c^2 R) \Delta v$. This balances the momentum given to the sphere. – John Eastmond Oct 14 '13 at 7:04
• You are right, the last part of my answer was incorrect before. The negative counter-momentum is not only in the sphere, but gets transported away with the acceleration field. I have edited the answer to account for this. Thanks. – Ján Lalinský Oct 30 '13 at 20:21