Q16P - Fundamentals Of Physics

Question

A parallel-plate capacitor with circular plates of radius $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{\text{m}}$ is being discharged. A circular loop of radius $\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{ }\mathit{m}$ is concentric with the capacitor and halfway between the plates. The displacement current through the loop is $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{A}$. At what rate is the electric field between the plates changing?

Step-by-Step Solution

Answer

The rate of the electric field between the plates changing is $7.2\times {10}^{12}V/m.s$

Given data

Displacement current, $i_d=2.0 \text{A}$ 

Circular plate radius, $R=0.10 \text{m}$

${\epsilon }_0=8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N.m}}^{\text{2}}$

 

Determining the concept

In a parallel plate capacitor, the real current $\mathit{i}$, which charges the plates, changes the electric field $\overrightarrow{\mathbf{E}}$ between the plates. The fictitious displacement current ${\mathit{i}}_{\mathbf{d}}$ between the plates is associated with that charging field, $\overrightarrow{\mathbf{E}}$. Therefore, we consider ${\mathbf{i}}_{\mathbf{d}}\mathbf{=}{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{A}\left(\mathbf{dE}\mathbf{/}\mathbf{dt}\right)$. In this problem, area A is the area over which a changing electric field is present. Also, it is given that, $\mathbf{r}\mathbf{>}\mathbf{R}$.

 

The formula is as follows:

 $i_d={\epsilon }_{0}A\left(dE/dt\right)$

where, $i_d$is the displacement current,$A$ is the area and,$E$ is the charging field.

Determining the rate of the electric field between the plates changing

To find the rate of electric field between plates with respect to the time,

$i_d={\epsilon }_{0}A\left(dE/dt\right)$

$$\begin{gathered} \frac{dE}{dt}=\frac{i_d}{{\epsilon }_{0}A} \\ =\frac{i_d}{{\epsilon }_0\pi R^2} \end{gathered}$$

Putting values,

$$\begin{gathered} \frac{dE}{dt}=\frac{2.0}{\left(8.85\times {10}^{-12}\right)\left(3.14\right){(0.10)}^2} \\ =7.2\times {10}^{12}\text{V/m.s} \end{gathered}$$

Therefore, the rate of the electric field between the plates changing is $7.2\times {10}^{12}\text{V/m.s}$.