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Q13P

Expert-verifiedFound in: Page 967

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${\mathbf{2}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathit{\mu}}{\mathbf{\text{F}}}$ capacitance be changed to produce a displacement current of ${\mathbf{1}}{\mathbf{.}}{\mathbf{5}}{\mathbf{}}{\mathit{A}}$****?**

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

** **

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

${i}_{d}=1.5\text{A}$

**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.

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,

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

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

** **

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