Plate capacitor: Maxwell stress tensor and forces

Question

I've got a plate capacitor with infinitely large plates at $z_1=d/2$ and $z_2=-d/2$. The plate at $z_1$ has a surface charge density of $\sigma$. The plate at $z_2$ has a surface charge density of $-\sigma$. First I calculated the electric field of the plate capacitor. Inside:$\vec{E} = - 4\pi \sigma \vec{e}_z$; Outside: $\vec{E} = 0$, The capacitor is inside a magnetic field $\vec{B} = B \vec{e}_x$.

After that I calculated the Maxwell stress tensor which is $$ T = \operatorname{diag}\left( \frac{B^2}{8\pi} - 2\pi \sigma^2, -\frac{B^2}{8\pi} - 2\pi \sigma^2, -\frac{B^2}{8\pi}+2\pi \sigma^2 \right). $$

Now I want to calculate the forces which act on the upper and lower plate. The general formula is $$\vec{F}_j = \int_S \vec{n} T_j $$ With $\vec{n} = \vec{e}_z$ for the upper plate and $\vec{n} = -\vec{e}_z$ for the lower plate, I get $\vec{F}_x = \vec{F}_y = 0$ which makes sense if I look at the symmetry here. If I want to calculate $\vec{F}_z$, I get an infinite force. Does that make sense?

Answer 1

You are getting infinite force because you are assuming infinite area. Real capacitors have finite area, and so they experience finite force. A way to overcome this is to calculate the force per unit area instead

$$\frac{F_{z}}{S}=\frac{1}{S}\int_{S}\hat{\rm T}\cdot{\rm}\vec{{\rm d}a}=2\pi\sigma^{2}-\frac{B^{2}}{8\pi}$$

which is finite. Lets calculate the $B=0$ case in another way to check our result. The potential energy of a capacitor is given by

$$V=\frac{Q^{2}}{2C}=\frac{\sigma^{2}S^{2}}{2\frac{S}{4\pi x}}=2\pi\sigma^{2}Sx$$

where $x$ is the distance between the plates. Thus the force per area on the plates is

$$\frac{F_{z}}{S}=\frac{1}{S}\frac{{\rm d}V}{{\rm d}x}=2\pi\sigma^{2}$$

exactly as before.

Answer 2

An infinite area under a finite presure are obviously under an infinite force $F=P\times A $. But no worry though, if the plate have a finite density of mass, the total mass will be infinite too. This make sense, infinite force for infinite inertia, witch can give a finite acceleration $a_c=F/M $.

Inifinity in physics almost all the times are just a limit, of something very big if you compare with typical sizes of the measurement apparatus.