Consider an infinite parallel-plate capacitor, with the lower plate (at $\mathit{z}\mathbf{=}\mathbf{-}\frac{\mathbf{d}}{\mathbf{2}}$ ) carrying surface charge density$\mathbf{-}\mathit{\sigma }$ , and the upper plate (at$\mathit{z}\mathbf{=}\mathbf{+}\frac{\mathbf{d}}{\mathbf{2}}$ ) carrying charge density $\mathbf{+}\mathit{\sigma }$.

(a) Determine all nine elements of the stress tensor, in the region between the plates. Display your answer as a $\mathbf{3}\mathbf{\times }\mathbf{3}$matrix:

$\left(\begin{array}{ccc}{\mathbf{T}}_{\mathbf{x}\mathbf{x}}& {\mathbf{T}}_{\mathbf{x}\mathbf{y}}& {\mathbf{T}}_{\mathbf{x}\mathbf{z}}\\ {\mathbf{T}}_{\mathbf{y}\mathbf{x}}& {\mathbf{T}}_{\mathbf{y}\mathbf{y}}& {\mathbf{T}}_{\mathbf{y}\mathbf{z}}\\ {\mathbf{T}}_{\mathbf{z}\mathbf{x}}& {\mathbf{T}}_{\mathbf{z}\mathbf{y}}& {\mathbf{T}}_{\mathbf{z}\mathbf{z}}\end{array}\right)$

(b) Use Eq. 8.21 to determine the electromagnetic force per unit area on the top plate. Compare Eq. 2.51.

(c) What is the electromagnetic momentum per unit area, per unit time, crossing the xy plane (or any other plane parallel to that one, between the plates)?

(d) Of course, there must be mechanical forces holding the plates apart—perhaps the capacitor is filled with insulating material under pressure. Suppose we suddenly remove the insulator; the momentum flux (c) is now absorbed by the plates, and they begin to move. Find the momentum per unit time delivered to the top plate (which is to say, the force acting on it) and compare your answer to (b). [Note: This is not an additional force, but rather an alternative way of calculating the same force—in (b) we got it from the force law, and in (d) we do it by conservation of momentum.]

Write the electric field in the x direction.

$E_x=0$

Write the electric field in the y-direction.

$E_y=0$

Write the electric field in the z-direction.

${\mathit{E}}_{\mathbf{z}}\mathbf{=}\mathbf{-}\frac{\mathbf{\sigma }}{{\mathbf{\epsilon }}_{\mathbf{0}}}$

(a)

Write the element in xx and yy direction.

${\mathrm{T}}_{\mathrm{xx}}={\mathrm{T}}_{\mathrm{yy}}=-\frac{{\mathrm{\epsilon }}_0}{2}{\mathrm{E}}^2$

Substitute the known values in the above equation.

$\begin{array}{c}{T}_{xx}=T_{yy}=-\frac{{\epsilon }_0}{2}{\left(-\frac{\sigma }{{\epsilon }_0}\right)}^2\\ T_{xx}=T_{yy}=-\frac{{\sigma }^2}{2{\epsilon }_0}\end{array}$

Write the element in the zz direction.

$T_{zz}={\epsilon }_0\left(E_z^2-\frac{1}{2}{E}^2\right)$

Substitute the known values in the above equation.

$\begin{array}{c}{T}_{zz}={\epsilon }_0\left[{\left(-\frac{\sigma }{{\epsilon }_0}\right)}^2-\frac{1}{2}{\left(-\frac{\sigma }{{\epsilon }_0}\right)}^2\right]\\ T_{zz}={\epsilon }_0\left[\frac{{\sigma }^2}{{\epsilon }_0^2}+\frac{{\sigma }^2}{2{\epsilon }_0^2}\right]\\ T_{zz}=\frac{{\sigma }^2}{2{\epsilon }_0}\end{array}$

Hence, all the nine elements of the stress tensor will be,

$\overleftrightarrow{T}=\frac{{\sigma }^2}{2{\epsilon }_0}\left(\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& +1\end{array}\right)$

Therefore, the nine elements of the stress tensor is$\overleftrightarrow{T}=\frac{{\sigma }^2}{2{\epsilon }_0}\left(\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& +1\end{array}\right)$ .

(b)

Write the expression for the electromagnetic force.

$F=\oint \overleftrightarrow{T}\cdot da$ …… (1)

Since $S=0$, and $B=0$, integrate over the xy plane. Hence,

$da=-dxdy\widehat{z}$

Here, a negative sign indicates the outward direction with respect to a surface enclosing the upper plate.

Substitute the known values in equation (1) for the z direction.

role="math" localid="1658236456305"

Hence, the electromagnetic force per unit area will be,

$f=\frac{F_{zz}}{A}=-\frac{{\sigma }^2}{2{\epsilon }_0}\widehat{z}$

Therefore, the electromagnetic force per unit area on the top plate is $-\frac{{\sigma }^2}{2{\epsilon }_0}\widehat{z}$.

(c)

The electromagnetic momentum per unit area, per unit time in the z-direction crossing a surface perpendicular to z will be,

$-T_{zz}=-\frac{{\sigma }^2}{2{\epsilon }_0}$

Therefore, the electromagnetic momentum per unit area, per unit time, crossing the xy plane is $-\frac{{\sigma }^2}{2{\epsilon }_0}$.

(d)

As the recoil force is the momentum delivered per unit time, the force per unit area on the top plate will be,

$f=-\frac{{\sigma }^2}{2{\epsilon }_0}\widehat{z}$

Hence, the result is the same as part (b).

Therefore, the momentum per unit time delivered to the top plate is$f=-\frac{{\sigma }^2}{2{\epsilon }_0}\widehat{z}$ .