Question

A parallel plate capacitor has rectangular plates of area $400{\text{cm}}^2$ and separated by $2\text{mm}$ with air as a medium. What is the magnitude of charge on each plate if the applied potential difference is $200\text{Volts}$?
(Take ${ϵ}_o=8.85\times {10}^{-12}\text{F/m}$)

• A. $1770.8\times {10}^{-9}\text{C}$
• B. $3.54\times {10}^{-6}\text{C}$
• C. $3.54\times {10}^{-8}\text{C}$
• D. $1770.8\times {10}^{-13}\text{C}$
  

Answer 1

The correct option is C $3.54\times {10}^{-8}\text{C}$

We can solve this problem in two simple steps.


Step $1:$

First, let's find the capacitance of the capacitor by using, $C=\frac{A{ϵ}_0}{d}$ we get,

$\Rightarrow C=\frac{400\times {10}^{-4}\times 8.85\times {10}^{-12}}{2\times {10}^{-3}}$

$\Rightarrow C=17.7\times {10}^{-11}\text{F}$

Step $2:$ Now Let's use $Q=CV$

$\Rightarrow Q=17.7\times {10}^{-11}\times 200=3.54\times {10}^{-8}\text{C}$

Hence, option $\left(b\right)$ is correct.