Electrostatics 5 Question 9
10. 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, $\varepsilon _0=8.85 \times 10^{-12} \frac{C^{2}}{N-m^{2}}$
(a) $9.85 \times 10^{-10} C$
(Main 2019, 12 Jan II)
(b) $8.85 \times 10^{-10} C$
(c) $7.85 \times 10^{-10} C$
(d) $6.85 \times 10^{-10} C$
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Answer:
Correct Answer: 10. $0.198 \mu A$
Solution:
- If $Q=$ charge on each plate, then
$$ Q=C V=\frac{\varepsilon _0 A}{d} \cdot E d=\varepsilon _0 A E $$
Here, $A=1 m^{2}, E=100 N / C$
and
$$ \varepsilon _0=8.85 \times 10^{-12} \frac{C^{2}}{N-m^{2}} $$
So, by substituting given values, we get $Q=8.85 \times 10^{-12} \times 1 \times 100=8.85 \times 10^{-10} C$
