Question

At what rate must the potential difference between the plates of a parallel-plate capacitor with a$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\mu }\mathbf{\text{F}}$ capacitance be changed to produce a displacement current of$\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathit{A}$?

Short answer

The rate of potential is $7.5\times {10}^5\text{V/s}$.

Step-by-step solution

Given data

$C=2.0\mu \text{F}$

$i_d=1.5\text{A}$

Determining the concept

By using the concept of displacement current and by using equation 32-10, write the equation for flux as a function of potential, and from that, find the rate of change of potential difference. The rate of change of potential difference across the plates of the capacitor is displacement current per unit capacitance.

Formulae are as follows:

$i_d={\epsilon }_0\frac{d{\varphi }_E}{dt}$

where, 'i' is current,$\varphi$ is the flux.

Determining the rate of change of potential difference

From the equation 32-10,

$i_d={\epsilon }_0\frac{d{\varphi }_E}{dt}$

Where, $\varphi =AE$theand the$E=\frac{V}{d}$.

So,

${\varphi }_E=\frac{AV}{d}$

So,

$i_d=\frac{A{\epsilon }_0}{d}\frac{dV}{dt}$

Where, the capacitance,

$C=\frac{A{\epsilon }_0}{d}$

So, the rate of potential is,

$$\begin{gathered} \frac{dV}{dt}=\frac{i_d}{C} \\ \frac{dV}{dt}=\frac{1.5}{2.0\times {10}^{-6}} \\ \frac{dV}{dt}=7.5\times {10}^5\text{V/s} \end{gathered}$$

Hence, the rate of potential is $7.5\times {10}^{5}V/s$.