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Found in: Page 967

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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

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

See the 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, $\varphi =AE$the and 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,

$\frac{dV}{dt}=\frac{{i}_{d}}{C}\phantom{\rule{0ex}{0ex}}\frac{dV}{dt}=\frac{1.5}{2.0×{10}^{-6}}\phantom{\rule{0ex}{0ex}}\frac{dV}{dt}=7.5×{10}^{5}\text{V/s}\phantom{\rule{0ex}{0ex}}$

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