Attractive Force of Spherical Capacitors

The reason is a bit subtle. It's fundamentally a result of the fact that the electric field is not continuous at the surface of the conducting shell.

Let's try the naive approach: the potential between the spherical capacitors is:

$V = \frac{Q}{4\pi \epsilon_0} \frac{b-a}{ab}.$

The electric field at radius $b$ due to this potential is:

$E = -\biggl.\frac{dV}{dr}\biggr|_b=-\biggl.\frac{d}{dr}\left(\frac{Q}{4\pi \epsilon_0} \frac{r-a}{ar}\right)\biggr|_b = -\frac{Q}{4\pi \epsilon_0 b^2},$

acting in the radial direction. It seems like our force will be off by a factor of $\frac12$ if we just multiply by a factor of $Q$ .

However, there are a couple problems. First of all, it is a little sketchy to say that the electric field on the interior of the outer sphere can see the outer sphere as a point charge at the center (which is basically what $F=qE$ says). Shell theorem holds outside a spherically symmetric charge distribution, not inside. Really to be fully rigorous here one ought to compute the force on a small element of the surface $dq = \sigma dA$ where $dA$ is an area element and $\sigma$ is the surface charge density. Then, since the electric field is a vector, you'll have to write down the components of the force in each direction and perform all the double integrals, and put them back together, and compute the magnitude. Of course, if you perform the integrals over the entire sphere every surface element has an antipodal element which will apparently cancel it in the integral! One way of dealing with this is by treating the sphere as two separate hemispheres and computing the force along the axis of each hemisphere. It's still ugly, and you'd have to compute the double integrals over each hemisphere anyway. So being rigorous about an $F=qE$ derivation is quite messy!

In any case, this isn't really where the missing factor of $\frac12$ comes from. This is actually derived from the fact that change in electric field on either side of a conductor is:

$\Delta E = \frac{\sigma}{\epsilon_0}$

where $\sigma$ is the surface charge density. One can show from this (see e.g. pages 31-32 of Purcell's Electricity and Magnetism , Third Edition, or try to derive it yourself!) that the force per unit area on a surface of charge is actually given by the average of the electric field on either side, not just the value inside the surface! Since the electric field is zero outside a capacitor (total charge is zero), this is where we get the extra factor of $\frac12$ , from the discontinuity in the electric field at the surface.

But the energy-of-electric-field argument circumvents all of these problems, which is why the original solution is much shorter (and I didn't even rigorously derive the solution from first principles in this response) :)

There is a slight abuse of notation in my solution in that I have used b both as the variable and the actual location of the outer shell. See my response to Mudit's question above where I write the force as the negative derivative with respect to r evaluated at b instead - the meaning is the same. The fact that force exerted on something in a given potential goes like the negative derivative of the potential follows for instance from the Lagrangian formalism of mechanics.