Q67 / 90 · JEE Main 2019 (12 Jan Shift 2) · Physics

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: $(Take \in_{0} = 8.85 \times 10^{- 12} \frac{C^{2}}{N - m^{2}})$

Correct answer
A
$8.85 x 10^{- 10} C$
B
$6.85 x 10^{- 10} C$
C
$9.85 x 10^{- 10} C$
D
$7.85 x 10^{- 10} C$

Solution

E = \frac{σ}{ε_{0}} = \frac{Q}{A ε_{0}} = 100 V / m

Q = 100 \times A ε_{0}

\Rightarrow Q = 100 \times 1 \times 8.85 \times 10^{- 12} C

= 8.85 \times 10^{- 10} C

Hence, the correct answer is option: 'A'.

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