91Ó°ÊÓ

Write the expression of Maxwell-stress tensor:

${\mathit{T}}_{\mathbf{i}\mathbf{j}}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\left(E_i{E}_j-\frac{1}{2}{\mathrm{δ}}_{ij}{E}^2\right)\mathbf{+}\frac{\mathbf{1}}{{\mathbf{μ}}_{\mathbf{0}}}\left(B_i{B}_j-\frac{1}{2}{\mathrm{δ}}_{ij}{B}^2\right)$…… (1)

Here, T_ijis the magnitude of the force acting per unit area in the i^th direction of the surface, which is oriented in the j^thdirection, E is the electric field, and B is the magnetic field.

(a)

Consider a plane equidistant from the two equal point charges.

Write the force on the upper charge.

$d{a}_x=d{a}_y=0$

Write the area vector in the z-direction.

$d{a}_z=-rdrd\mathrm{Ï•}$

Write the net force in the z-direction.

$$\begin{gathered} F=\stackrel{→}{T}·d{a}_z \\ F=\left(T_{zx}d{a}_x+T_{zy}d{a}_y+T_{zz}d{a}_z\right) \end{gathered}$$

Using equation (1), the Maxwell-stress tensor equation becomes,

${\left(\stackrel{→}{T}·da\right)}_z={\mathrm{Î\mu }}_0\left(E_z{E}_z-\frac{1}{2}{E}^2\right)\left(-rdrd\mathrm{Ï•}\right)$…… (2)

Write the electric field due to a point charge.

$E=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q}{R^2}\hat{x}$

From the above figure, resolve the electric field due to both charges into components and write the resultant field along the x-axis.

$E=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{2q}{R^2}\cos \mathrm{θ}\hat{x}$…… (3)

As the component $E\sin \mathrm{θ}$is in the opposite direction, the electric field will be zero.

From the above figure, the data is observed as:

$\begin{array}{rcl}\cos \mathrm{θ}& =& \frac{a}{R}\left(∴R=\sqrt{x^2+a^2}\right)\\ \hat{r}& =& \frac{\stackrel{→}{r}}{\left|r\right|}\end{array}$

Substitute the value localid="1653736172255" $\cos \mathrm{θ}=\frac{a}{R}$of in equation (3).

localid="1653736183584" $\begin{array}{l}E=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{2q}{R^2}\left(\frac{a}{R}\right)\widehat{x}\\ E^2={\left[\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{2q}{R^2}\left(\frac{a}{R}\right)\right]}^2\\ E^2={\left(\frac{q}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)}^2{\left(\frac{a}{R^3}\right)}^2\\ E^2={\left(\frac{q}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)}^2\frac{a^2}{{\left(x^2+a^2\right)}^3}\end{array}$

Substitute the values in equation (2) and integrate it with the respective limits.

localid="1653736199693" $\begin{array}{c}F=∫{\left(\stackrel{→}{T}â‹\dots da\right)}_z\\ F=\underset{0}{\overset{2\mathrm{Ï€}}{∫}}{∫}_0^{\mathrm{∞}}\left(−\frac{1}{2}{\left(\frac{q}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)}^2\left(\frac{r^2}{{\left(r^2+a^2\right)}^3}\right)\right)−rdrd\mathrm{Ï•}\\ F=\frac{1}{2}{\left(\frac{q}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)}^2\left(2\mathrm{Ï€}\right){∫}_0^{\mathrm{∞}}\frac{r^{3}dr}{{\left(r^2+a^2\right)}^3}\\ F=\frac{q^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{∫}_0^{\mathrm{∞}}\frac{r^{3}dr}{{\left(r^2+a^2\right)}^3}\text{ }\mathrm{......}\text{ }\left(4\right)\end{array}$

Let’s assume,

localid="1653736209810" $\begin{array}{c}{r}^2=u\\ 2rdr=du\\ dr=\frac{du}{2r}\end{array}$

Substitute the value of rand drin equation (4).

localid="1653736221033" $\begin{array}{c}{F}_z=\frac{q^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{∫}_0^{\mathrm{∞}}\frac{u^3\left(\frac{du}{2r}\right)}{{\left(u^2+a^2\right)}^3}\\ F_z=\frac{q^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{2}{∫}_0^{\mathrm{∞}}\frac{udu}{{\left(u+a^2\right)}^3}\end{array}$

Using the standard integration method, calculate the net force acting on the top of the sheet.

localid="1653736235824" $\begin{array}{c}{F}_z=\frac{q^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{1}{2}\right){\left[−\frac{1}{\left(u+a^2\right)}+\frac{a^2}{2{\left(u+a^2\right)}^3}\right]}_0^{\mathrm{∞}}\\ F_z=\frac{q^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{1}{2}\right){\left[\left(0\right)−\left(\frac{1}{a^2}−\frac{a^2}{a^4}\right)\right]}_0^{\mathrm{∞}}\\ F_z=\frac{q^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{1}{2}\right)\left[0+\frac{1}{a^2}−\frac{a^2}{2a^4}\right]\\ F_z=\frac{q^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{{\left(2a\right)}^2}\end{array}$

(b)

Consider a plane equidistant from the two equal and unlike point charges.

From the above figure, resolve the electric field due to both equal and unlike charges into components and write the resultant electric field.

$E=−\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{2q}{r^2}\sin \mathrm{θ}\widehat{z}$

Substitute the value of $\sin \mathrm{θ}=\frac{a}{r}$in the above expression.

$\begin{array}{l}E=−\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{2q}{r^2}\sin \mathrm{θ}\widehat{z}\\ E^2={\left[\left(\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{2q}{r^2}\right)\left(\frac{a}{r}\right)\right]}^2\\ E^2={\left(\frac{qa}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)}^2{\left(\frac{r}{r^3}\right)}^2\\ E^2={\left(\frac{qa}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)}^2\left(\frac{1}{{\left(r^2+a^2\right)}^3}\right)\end{array}$

Substitute the values in equation (2) and integrate it with the respective limits.

$\begin{array}{c}F=∫{\left(\stackrel{→}{T}â‹\dots da\right)}_z\\ F=∫−{\mathrm{Î\mu }}_0\left(E_z^2−\frac{1}{2}{E}^2\right)\left(−rdrd\mathrm{Ï•}\right)\\ F=∫−\frac{{\mathrm{Î\mu }}_0}{2}{\left(\frac{qa}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)}^2\frac{1}{{\left(r^2+a^2\right)}^3}rdrd\mathrm{Ï•}\\ F=−\frac{{\mathrm{Î\mu }}_0}{2}{\left(\frac{qa}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)}^2\underset{0}{\overset{2\mathrm{Ï€}}{∫}}\underset{0}{\overset{\mathrm{∞}}{∫}}\frac{rdrd\mathrm{Ï•}}{{\left(r^2+a^2\right)}^3}\end{array}$

On further solving,

$\begin{array}{c}F=−\frac{q^2{a}^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{∫}_0^{\mathrm{∞}}\frac{rdr}{{\left(r^2+a^2\right)}^3}\\ F=−\frac{q^2{a}^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left[−\frac{1}{4}\frac{1}{{\left(r^2+a^2\right)}^2}\right]}_0^{\mathrm{∞}}\\ F=−\frac{q^2{a}^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left[0+\frac{1}{4a^4}\right]\\ F=−\frac{q^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{{\left(2a\right)}^2}\end{array}$

Therefore, the force of one charge on the other is $F=−\frac{q^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{{\left(2a\right)}^2}$.