Chapter 30, Problem 38P

(a)

To determine

The proof that the displacement current in the capacitor gap has the same value as the conduction current in the capacitor leads.

Answer to Problem 38P

The displacement current in the capacitor gap has the same value as the conduction current in the capacitor leads/

Explanation of Solution

Given:

The radius of the circular plateis $b$ .

The separation distance between the plates is $d$

Formula used:

The expression for displacement current is given by,

  $I_d={\epsilon }_0\frac{d{\varphi }_c}{dt}$

The expression for electric flux is given by,

  ${\varphi }_c=EA$

The expression for electric field strength between the plates of the capacitor is given by,

  $E=\frac{Q}{{\epsilon }_{0}A}$

Calculation:

The displacement current is evaluated as,

  $\begin{array}{c}{I}_d={\epsilon }_{0}A\frac{dE}{dt}\\ ={\epsilon }_{0}A\frac{d\left(\frac{Q}{{\epsilon }_{0}A}\right)}{dt}\\ ={\epsilon }_{0}A\left(\frac{1}{{\epsilon }_{0}A}\right)\frac{dQ}{dt}\\ =\frac{dQ}{dt}\end{array}$

Solve further as,

  $I_d=I$

Here, $I=\frac{dQ}{dt}$ is the conduction current in capacitor leads.

Conclusion:

Therefore, the displacement current in the capacitor gap has the same value as the conduction current in the capacitor leads.

(b)

To determine

The direction of pointing vector in the region between the capacitor plates.

Explanation of Solution

Introduction:

The Poynting vector is a quantity which describes the magnitude and direction of the flow of energy in electromagnetic waves. It is mathematicallyrepresented as,

  $\overrightarrow{S}=\frac{1}{{\mu }_0}\left(\overrightarrow{E}\times \overrightarrow{B}\right)$

The electric field $\overrightarrow{E}$ is always perpendicular to the plates of the capacitor and the magnetic field $\overrightarrow{B}$ is tangent to the concentric circles whose center is through the middle of the capacitor plates and as pointing vector $\overrightarrow{S}$ is perpendicular to the plane containing $\overrightarrow{E}$ and $\overrightarrow{B}$ .

Using the equation of pointing vector the direction of $\overrightarrow{S}$ points radially inward that is towards the center of the capacitor.

Conclusion:

Therefore, the direction of pointing vector is radially inward.

(c)

To determine

The expression for poynting vector in the region between the capacitor plates and to prove that flux into the region between the plates is equal to the rate of change of the energy stored in the capacitor.

Answer to Problem 38P

The expression for poynting vector in the region between the capacitor plates is $-\frac{{\epsilon }_{0}E}{2}\frac{dE}{dt}R\widehat{R}$

Explanation of Solution

Formula used:

The expression for Ampere’s Law is given by,

  ${\displaystyle \underset{c}{\oint }\overrightarrow{B}}\cdot \overrightarrow{dl}={\mu }_{0}I$

The expression for total energy stored in the capacitor is given by,

  $U=uV$

Calculation:

The direction of $\overrightarrow{E}$ is in the $+x$ direction.

The diagrammatical representation is shown below.

  

Figure 1

Applying Ampere’s law to a closed circular path of radius $R\le b$ ,

  $\begin{array}{c}B\left(2\pi R\right)={\mu }_0{I}_d\\ B\left(2\pi R\right)={\mu }_0{\epsilon }_0\frac{d}{dt}EA\\ B\left(2\pi R\right)={\mu }_0{\epsilon }_0\pi R^2\frac{dE}{dt}\\ B=\frac{{\mu }_0{\epsilon }_{0}R}{2}\frac{dE}{dt}\end{array}$

In vector notation

  $\overrightarrow{B}=\frac{{\mu }_0{\epsilon }_{0}R}{2}\frac{dE}{dt}\widehat{j}$

Substituting $\overrightarrow{E}$ and $\overrightarrow{B}$ in the equation of pointing vector.

  $\begin{array}{l}\overrightarrow{S}=\left(\frac{1}{{\mu }_0}E\right)\widehat{i}\times \left(-\frac{{\mu }_0{\epsilon }_0}{2}R\frac{dE}{dt}\right)\widehat{j}\\ =\frac{{\epsilon }_{0}E}{2}\frac{dE}{dt}R\left(\widehat{i}\times \left(-\widehat{j}\right)\right)\\ =-\frac{{\epsilon }_{0}E}{2}\frac{dE}{dt}R\widehat{R}\end{array}$

Here, $\widehat{R}$ is a unit vector pointing radially outward from the line joining the centers of the plates.

The total energy in the capacitor is calculated as.

  $U=\left({\epsilon }_0{E}^2\right)\left(Ad\right)$

The rate at which energy is stored in the capacitor is calculated as,

  $\begin{array}{c}\frac{dU}{dt}=\frac{d}{dt}\left(uV\right)\\ =V\frac{d}{dt}\left({\epsilon }_0{E}^2\right)\\ =Ad{\epsilon }_{0}E\frac{dE}{dt}\end{array}$

Further solve as,

  $\begin{array}{c}\frac{dU}{dt}=Ad{\epsilon }_0\left(\frac{Q}{{\epsilon }_{0}A}\right)\frac{d}{dt}\left(\frac{Q}{{\epsilon }_{0}A}\right)\\ =\frac{Qd}{{\epsilon }_{0}A}\frac{dQ}{dt}\\ =\frac{Qd}{{\epsilon }_{0}A}I\end{array}$

The energy flowing into the solenoid per unit time is written as,

  $\begin{array}{c}{\displaystyle \oint S\cdot dA}=S\left(2\pi bd\right)\\ =\left(\frac{1}{2}{\epsilon }_{0}Eb\frac{dE}{dt}\right)\left(2\pi bd\right)\\ =\pi {\epsilon }_{0}E{b}^{2}d\frac{dE}{dt}\\ =\pi {\epsilon }_0\left(\frac{Q}{{\epsilon }_{0}A}\right)b^{2}d\frac{d}{dt}\left(\frac{Q}{{\epsilon }_{0}A}\right)\end{array}$

Further solve as,

  $\begin{array}{c}{\displaystyle \oint S\cdot dA}=\pi \left(\frac{Q}{\pi b^2}\right)\frac{b^{2}d}{{\epsilon }_{0}A}\frac{dQ}{dt}\\ =\frac{Qd}{{\epsilon }_{0}A}I\end{array}$

So, the flux into the region between the plates is equal to the rate of change of the energy stored in the capacitor

Conclusion:

Therefore, the expression for poynting vector in the region between the capacitor plates is $-\frac{{\epsilon }_{0}E}{2}\frac{dE}{dt}R\widehat{R}$