JEE PYQ: Capacitance Question 41
Question 41 - 2019 (12 Jan Shift 1)
A parallel plate capacitor with plates of area 1 m$^2$ each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is:
(Take $\epsilon_0 = 8.85 \times 10^{-12}$ $\frac{C^2}{N \cdot M^2}$)
(1) $7.85 \times 10^{-10}$ C
(2) $6.85 \times 10^{-10}$ C
(3) $8.85 \times 10^{-10}$ C
(4) $9.85 \times 10^{-10}$ C
Show Answer
Answer: (3)
Solution
$E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}$. $Q = \epsilon_0 EA = 8.85 \times 10^{-12} \times 100 \times 1 = 8.85 \times 10^{-10}$ C.
BETA
