Question

A parallel plate capacitor with plates of area $1{\text{m}}^2$ each, are at a separation $0.1\text{m}$. If the electric field between the plates is $100{\text{NC}}^{-1}$, the magnitude of charge on each plate is:
$(\text{Take}{\epsilon }_0=8.85\times {10}^{-12}\frac{{\text{C}}^2}{{\text{Nm}}^2})$

• A. $8.85\times {10}^{-10}\text{C}$
• B. $7.85\times {10}^{-10}\text{C}$
• C. $6.85\times {10}^{-10}\text{C}$
• D. $9.85\times {10}^{-10}\text{C}$

Answer 1

The correct option is A $8.85\times {10}^{-10}\text{C}$

The electric field between the plates of a parallel plate capacitor is given by,

$E=\frac{\sigma }{{\epsilon }_0}=\frac{Q}{A{\epsilon }_0}$
$\Rightarrow Q={\epsilon }_0.E.A=8.85\times {10}^{-12}\times 100\times 1$
$\Rightarrow Q=8.85\times {10}^{-10}\text{C}$

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