Question

A capacitor made of two circular plates each of radius $12\text{cm}$ and separated by $5\text{cm}$. The capacitor is being charged by an external source. The charging current is constant and is equal to $0.15\text{A}$. The rate of change of potential difference between the plates is

• A. $8.39\times {10}^{10}\text{V/s}$
• B. $1.65\times {10}^{11}\text{V/s}$
• C. $2.34\times {10}^{10}\text{V/s}$
• D. $1.87\times {10}^{10}\text{V/s}$

Answer 1

The correct option is D $1.87\times {10}^{10}\text{V/s}$

Given :-
Radius of each plate $\left(r\right)=12\text{cm}$
distance between the plates $\left(d\right)=5\text{cm}$
Conduction current $\left(I_c\right)=0.15\text{A}$

The capacitance is given by :-

$C=\frac{{\epsilon }_{0}A}{d}=\frac{{\epsilon }_0\pi r^2}{d}$

$C=\frac{8.85\times {10}^{-12}\times \pi \times (12\times {10}^{-2}{)}^2}{5\times {10}^{-2}}$

$\therefore C=8\times {10}^{-12}\text{F}$

Charge on each plate of the capacitor is $q=CV$

Where, $V$ is the potential difference between the plates

$\therefore I_c=\frac{dq}{dt}=\frac{d}{dt}\left(CV\right)=C\frac{dV}{dt}$

$\frac{dV}{dt}=\frac{I_c}{C}=\frac{0.15}{8\times {10}^{-12}}$

$\therefore \frac{dV}{dt}=1.87\times {10}^{10}\text{V/s}$

Hence, option $\left(\text{D}\right)$ is correct.