Imagine two parallel infinite sheets, carrying uniform surface charge $+\sigma$(on the sheet at $z=d$) and $-\sigma$(at $z=0$). They are moving in they direction at constant speed v (as in Problem 5.17).

(a) What is the electromagnetic momentum in a region of area A?

(b) Now suppose the top sheet moves slowly down (speed u) until it reaches the bottom sheet, so the fields disappear. By calculating the total force on the charge $\left(q=\sigma A\right)$, show that the impulse delivered to the sheet is equal to the momentum originally stored in the fields. [Hint: As the upper plate passes by, the magnetic field drops to zero, inducing an electric field that delivers an impulse to the lower plate.]

Write the expression for the electromagnetic momentum density.

$g_{em}={\epsilon }_0\left(E\times B\right)$ …… (1)

Here, E is the electric field, and B is the magnetic field.

(a)

Write the expression for the electric field.

$E=-\frac{\sigma }{{\epsilon }_0}\widehat{z}$

Write the expression for the magnetic field.

$B=-{\mu }_0\sigma v\widehat{x}$

Substitute all the known values in equation (1).

$\begin{array}{c}{g}_{\text{em}}={\epsilon }_0\left(-\frac{\sigma }{{\epsilon }_0}\widehat{z}\times \left(-{\mu }_0\sigma v\widehat{x}\right)\right)\\ g_{\text{em}}={\mu }_0{\sigma }^{2}v\widehat{y}\end{array}$

Write the expression for the electromagnetic momentum in a region of area A.

$p_{\text{em}}=\left(dA\right)g_{\text{em}}$

Substitute the known values in the above expression.

$\begin{array}{c}{p}_{\text{em}}=\left(dA\right)\left(-{\mu }_0{\sigma }^{2}v\widehat{y}\right)\\ p_{\text{em}}={\mu }_0{\sigma }^{2}vdA\widehat{y}\end{array}$

Therefore, the electromagnetic momentum in a region of area A is $p_{\text{em}}={\mu }_0{\sigma }^{2}vdA\widehat{y}$.

(b)

Write the expression for the initial impulse delivered to the system.

$I_1=\int F\cdot dt$ ….. (2)

Here, F is the magnetic force which is given as:

Determine the expression for the magnetic force.

$\begin{array}{c}F=q\left(u\times B\right)\\ F=\sigma A\left[\left(-u\widehat{z}\right)\times \left(-\frac{1}{2}{\mu }_0\sigma v\widehat{x}\right)\right]\\ F=\frac{1}{2}{\mu }_0{\sigma }^{2}Avu\widehat{y}\end{array}$

Substitute the known value in equation (2).

$\begin{array}{c}{I}_1=\int \frac{1}{2}{\mu }_0{\sigma }^{2}Avu\widehat{y}dt\\ I_1=\frac{1}{2}{\mu }_0{\sigma }^{2}Av\widehat{y}\int udt\\ I_1=\frac{1}{2}d{\mu }_0{\sigma }^{2}Av\widehat{y}\end{array}$

Write the expression for the final impulse delivered to the system.

$I_2=\int F\cdot dt$ ….. (3)

Here, F is the magnetic force which is given as:

$\begin{array}{c}F=q\left(u\times B\right)\\ F=\sigma A\left[\left(-u\widehat{z}\right)\times \left(-\frac{1}{2}{\mu }_0\sigma v\widehat{x}\right)\right]\\ F=\frac{1}{2}{\mu }_0{\sigma }^{2}Avu\widehat{y}\end{array}$

Substitute the known value in equation (3).

$\begin{array}{c}{I}_2=\int \frac{1}{2}{\mu }_0{\sigma }^{2}Avu\widehat{y}dt\\ I_2=\frac{1}{2}{\mu }_0{\sigma }^{2}Av\widehat{y}\int udt\\ I_2=\frac{1}{2}d{\mu }_0{\sigma }^{2}Av\widehat{y}\end{array}$

Hence, the total impulse delivered to the sheet will be,

$\begin{array}{c}I=I_1+I_2\\ I=\left(\frac{1}{2}d{\mu }_0{\sigma }^{2}Av\widehat{y}\right)+\left(\frac{1}{2}d{\mu }_0{\sigma }^{2}Av\widehat{y}\right)\\ I=dA{\mu }_0{\sigma }^{2}v\widehat{y}\end{array}$

Hence proved.

Therefore, the impulse delivered to the sheet is equal to the momentum originally stored in the fields.