91Ó°ÊÓ

(a) Edge length of square plate of the parallel plate capacitor, $L=1.0\text{m}$

(b) Current charging the capacitor, $I=2.0\text{A}$

(c) Current produces a uniform electric field perpendicular to the plates.

(d) Edge length of the dashed square path,$d=0.05\text{m}$

When two conducting plates are arranged parallel insulated to each other with a gap filled with a dielectric material, they act as parallel-plate capacitors. The plates conduct the flow of current such that it stores some amount of charge within their dielectric space, thus, they act as electrodes. The capacitor is constructed with the view of storing some amount of electric energy that is further used to provide the circuit with the required amount of energy when needed.

Formulae:

The amount of current flow with a circuit, $i=\frac{dq}{dt}$ (i)

where, $q$is amount of the charge flowing through the material, and is time of charging.

The magnetic flux equation according to Ampere’s law, $âˆ\circledR B.ds={\mathrm{μ}}_{0}i$ (ii)

where, $i$is current, $B$is magnetic induction, $ds$is the area of the enclosed region ${\mathrm{μ}}_0=\left(4\mathrm{π}×{10}^{-7}\text{T.m/A}\right)$is the magnetic permittivity constant.

The amount of charge stored within the capacitor, $q=CV$ (iii)

where, $C$is the capacitance of the capacitor, $V$is the potential difference between the plates.

The capacitance of a capacitor,$C=\frac{{\mathrm{Î\mu }}_{0}A}{d}$ (iv)

where, ${\mathrm{Î\mu }}_0=\left(8.85×{10}^{-12}{\text{C}}^2/{\text{N.m}}^2\right)$is the permittivity in vacuum, $A$ is the area of the plates, $d$is the separation between the plates.

The electric field between the plates in a parallel-plate capacitor is changing, so there

is a nonzero displacement current between the plates.

Let,$A$be the area of a plate and E be the magnitude of the electric field between the plates.

The field between the plates is uniform, so the magnitude of the electric field can be given as:$E=V/d$

where, $V$is the potential difference across the plates, and $d$is the plate separation. The current flowing into the positive plate of the capacitor can be given by substituting the charge value of equation (iii) and capacitance value of equation (iv) in equation (i) as follows:

$$\begin{gathered} i=\frac{d\left(CV\right)}{dt} \\ =\frac{C·dV}{dt} \\ =\left(\frac{{\mathrm{Î\mu }}_{0}A}{d}\right)\frac{d\left(Ed\right)}{dt}\left(âˆ\mu E=V/d\text{and using equation (iv)}\right) \\ =\frac{{\mathrm{Î\mu }}_{0}AdE}{dt} \\ =\frac{{\mathrm{Î\mu }}_{0}d{\mathrm{Ï•}}_E}{dt}\left(âˆ\mu \text{Electric flux in the region,}{\mathrm{Ï•}}_E=EA\right).....................\left(I\right) \end{gathered}$$

which has the same value as the displacement current formula, that is given as:

The displacement current through the wire, $i_d={\mathrm{Î\mu }}_0\frac{d{\mathrm{Ï•}}_E}{dt}$

where, ${\mathrm{Î\mu }}_0=\left(8.85×{10}^{-12}{\text{C}}^2/{\text{N.m}}^2\right)$is the permittivity in vacuum, is the electric flux through the region.

At any instant, the displacement current in the gap between the plates equals the

conduction current i in the wires. Thus, the value of the displacement current will be

$$\begin{gathered} i_d=i \\ =2.0\text{A} \end{gathered}$$

The rate of change of the electric field can be given using equation (I) in part (a) calculations as follows:

$$\begin{gathered} \frac{dE}{dt}=\frac{i_d}{{\mathrm{Î\mu }}_{0}A} \\ =\frac{2.0\text{A}}{\left(8.85×{10}^{-12}{\text{C}}^2/{\text{N.m}}^2\right){\left(1.0\text{m}\right)}^2}..........................\left(II\right) \\ =2.3×{10}^{11}\text{V/m}·\text{s} \end{gathered}$$

Hence, the rate of change of the electric field is $2.3×{10}^{11}\text{V/m}·\text{s}$.

Using equation (II) of part (b), the rate of electric field in the region for the given square with edge lengthcan be given as follows:

$\frac{d{E}_1}{dt}=\frac{i_d}{{\mathrm{Î\mu }}_0{L}^2}\left(âˆ\mu \text{Area of the square,}A=L^2\right)..........................\left(III\right)$

Using equation (II) of part (b), the rate of electric field in the region for the dashed square with edge lengthcan be given as follows:

$\frac{d{E}_2}{dt}=\frac{i_d\text{'}}{{\mathrm{Î\mu }}_0{d}^2}\left(âˆ\mu \text{Area of the dashed square,}A=d^2\right)..............\left(IV\right)$

Since, the electric field value is uniform in the region, using equation (III) and (IV), the value of the displacement current through the indicated or the dashed square path can be given as follows:

$$\begin{gathered} \frac{d{E}_1}{dt}=\frac{d{E}_2}{dt} \\ \frac{i_d}{{\mathrm{Î\mu }}_0{L}^2}=\frac{i_d\text{'}}{{\mathrm{Î\mu }}_0{d}^2} \\ {i}_d\text{'}=i_d\left(\frac{d^2}{L^2}\right) \\ =\left(2.0\text{A}\right){\left(\frac{0.50\text{m}}{1.0\text{m}}\right)}^2 \\ =0.50\text{A} \end{gathered}$$

Hence, the displacement current encircled by the square dashed path is $0.50\text{A}$.

The integral of the field around the indicated path is given using the given data in equation (ii) as follows:

$$\begin{gathered} âˆ\circledR B.ds={\mathrm{μ}}_0{i}_d \\ =\left(4\mathrm{Ï€}×{10}^{-7}\text{T.m/A}\right)\left(0.50\text{A}\right) \\ =63×{10}^{-7}\text{T}·\text{m} \end{gathered}$$

Hence, the integral of the field around the indicated path is $63×{10}^{-7}\text{T}·\text{m}$.