Question

Figure represents a capacitor made of two circular plates each of radius $r=12\text{cm}$ and separated by $d=5.0\text{mm}$. The capacitor is being charged by an external source. The charging current is constant current, $I=0.15\text{A}$.

Calculate the rate of change of electric field between the plates.

• A. $3.14\times {10}^{11}V{m}^{-1}{s}^{-1}$
• B. $2.68\times {10}^{11}V{m}^{-1}{s}^{-1}$
• C. $1.87\times {10}^{11}V{m}^{-1}{s}^{-1}$
• D. $5.12\times {10}^{11}V{m}^{-1}{s}^{-1}$

Answer 1

The correct option is A $3.14\times {10}^{11}V{m}^{-1}{s}^{-1}$

We know that
$Q=CV$
$\Rightarrow \frac{dQ}{dt}=C\frac{dV}{dt}$
As we know that
$V=E.d$
where ${}^{\prime }{d}^{\prime }$ is the distance between the plates
Putting this in above equation
$\Rightarrow \frac{dQ}{dt}=Cd\frac{dE}{dt}\Rightarrow I=C\frac{dE}{dt}d\Rightarrow \frac{dE}{dt}=\frac{I}{Cd}\Rightarrow \frac{dE}{dt}=\frac{I}{\frac{{\epsilon }_{0}A}{d}d}\Rightarrow \frac{dE}{dt}=\frac{I}{{\epsilon }_{0}A}\Rightarrow \frac{dE}{dt}=\frac{0.15}{8.85\times {10}^{-12}\times \pi (12\times {10}^{-2}{)}^2}\Rightarrow \frac{dE}{dt}=3.14\times {10}^{11}{\text{Vm}}^{-1}{\text{s}}^{-1}$