Question:
A parallel plate capacitor with plates of area $1 \mathrm{~m}^{2}$ each, are at a separation of $0.1 \mathrm{~m}$. If the electric field between the plates is $100 \mathrm{~N} / \mathrm{C}$, the magnitude of charge on each plate is:
(Take $\epsilon_{0}=8.85 \times 10^{-12} \frac{\mathrm{C}^{2}}{\mathrm{~N}-\mathrm{M}^{2}}$ )
Correct Option: , 3
Solution:
(3) $\mathrm{E}=\frac{\sigma}{\varepsilon_{0}}=\frac{\mathrm{Q}}{\mathrm{A} \varepsilon_{0}}$
$\therefore \mathrm{Q}=\varepsilon_{0} . \mathrm{E} . \mathrm{A}=8.85 \times 10^{-12} \times 100 \times 1$
$=8.85 \times 10^{-10} \mathrm{C}$