Consider the charging capacitor in Prob. 7.34.

(a) Find the electric and magnetic fields in the gap, as functions of the distance s from the axis and the timet. (Assume the charge is zero at $\mathit{t}\mathbf{=}\mathbf{0}$).

(b) Find the energy density ${\mathit{u}}_{\mathbf{e}\mathbf{m}}$and the Poynting vector S in the gap. Note especially the direction of S. Check that $\mathit{E}\mathit{q}\mathbf{.}\mathbf{8}\mathbf{.}\mathbf{12}$is satisfied.

(c) Determine the total energy in the gap, as a function of time. Calculate the total power flowing into the gap, by integrating the Poynting vector over the appropriate surface. Check that the power input is equal to the rate of increase of energy in the gap (Eq 8.9—in this case W = 0, because there is no charge in the gap). [If you’re worried about the fringing fields, do it for a volume of radius $\mathit{b}\mathbf{<}\mathit{a}$well inside the gap.]

Write the expression of Maxwell-Ampere Law:

$\oint B\cdot ds={\mu }_{0}I+{\mu }_0{\epsilon }_0\frac{d{\varphi }_E}{dt}$….. (1)

Here, B is the magnetic field produced by the moving charges, I is the current due to moving charges, is the permeability of free space and $\frac{d{\varphi }_E}{dt}$is the change in electric flux due to the change in velocity of the charged particles.

(a)

For closed curved paths, the integration of the magnetic field is zero. Hence, equation (1) becomes,

$\begin{array}{l}\oint B\cdot ds={\mu }_{0}I+{\mu }_0{\epsilon }_0\frac{d{\varphi }_E}{dt}\\ 0={\mu }_{0}I+{\mu }_0{\epsilon }_0\frac{d{\varphi }_E}{dt}\end{array}$

Rearrange the above equation for I.

$I=-{\epsilon }_0\frac{d{\varphi }_E}{dt}$….. (2)

Here, a negative sign indicates the direction of a current.

Write the formula for the electric flux.

${\varphi }_E=E\cdot A$….. (3)

Here, A is the area of cross section.

Write the formula for the area of cross-section of the plates of the capacitor.

$A=\pi r^2$….. (4)

Substitute the value of equations (3) and (4) in equation (2).

$\begin{array}{l}I={\epsilon }_0\frac{d\left(EA\right)}{dt}\\ I={\epsilon }_0\left(\pi r^2\right)\frac{dE}{dt}\\ dE=\frac{I}{{\epsilon }_0\left(\pi r^2\right)}dt\end{array}$

Integrate the above equation.

$\begin{array}{l}\int dE=\int \frac{I}{{\epsilon }_0\left(\pi r^2\right)}dt\\ E=\frac{It}{{\epsilon }_0\left(\pi a^2\right)}\\ \overrightarrow{E}=\frac{It}{\pi {\epsilon }_0{a}^2}\widehat{z}\end{array}$

Write the formula for the electric flux through the wire.

${\varphi }_E=\oint E\cdot da$….. (5)

As there is no current enclosed in the gap between the wires, the value of Iwill be zero.

Substitute the value of equation (5) in equation (1).

$\begin{array}{l}\oint B\cdot dl={\mu }_0{I}_{\text{enc}}+{\mu }_0{\epsilon }_0\frac{d{\varphi }_E}{dt}\\ \oint B\cdot dl={\mu }_0\left(0\right)+{\mu }_0{\epsilon }_0\frac{d\left(E\cdot da\right)}{dt}\\ B\left(2\pi s\right)=0+{\mu }_0{\epsilon }_0\left(\frac{dE}{dt}\right)\left(\pi s^2\right)\\ B\left(2\pi s\right)={\mu }_0{\epsilon }_0\left(\frac{I}{\pi {\epsilon }_0{a}^2}\right)\left(\pi s^2\right)\end{array}$

On further solving,

$\overrightarrow{B}=\frac{{\mu }_{0}Is}{2\pi a^2}\widehat{\varphi }$

Therefore, the electric and magnetic fields in the gap is $\overrightarrow{E}=\frac{It}{\pi {\epsilon }_0{a}^2}\widehat{z}$and $\overrightarrow{B}=\frac{{\mu }_{0}Is}{2\pi a^2}\widehat{\varphi }a$respectively.