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The figure shows a capacitor made of two circular plates each of radius 12 \mathrm{~cm} and separated by $5.0 \mathrm{~cm}$. The capacitor is being charged by an external source. The charging current is 0.15 \mathrm{~A}.

The rate of change of potential difference between the plates is:

Option: 1

1.87 \times 10^{10} \mathrm{~V} / \mathrm{s}


Option: 2

3.74 \times 10^{10} \mathrm{~V} / \mathrm{s}


Option: 3

7.50 \times 10^{10} \mathrm{~V} / \mathrm{s}


Option: 4

0.94 \times 10^{10} \mathrm{~V} / \mathrm{s}


Answers (1)

best_answer

\begin{aligned} c & =\frac{\epsilon_0 A}{d}=\frac{\epsilon_0\left(\pi R^2\right)}{d}=\frac{8.85 \times 10^{-12} \times 3.14 \times(0.12)^2}{5 \times 10^{-2}} \\ & =8 \times 10^{-12} \mathrm{~F}=8 \mathrm{PF} . \end{aligned}

Potential difference v=\frac{q}{c}.

\Rightarrow \frac{d V}{d t}=\frac{1}{c} \cdot \frac{d q}{d t}=\frac{i}{c}=\frac{0.15}{8 \times 10^{-12}}=1.87 \times 10^{10} \mathrm{~V} / \mathrm{s} .

Posted by

HARSH KANKARIA

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