Electrostatics Ques 144
- 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, $\varepsilon_{0}=8.85 \times 10^{-12} \frac{\mathrm{C}^{2}}{\mathrm{~N}-\mathrm{m}^{2}}\Big)$
(Main 2019, 12 Jan II)
(a) $9.85 \times 10^{-10} \mathrm{C}$
(b) $8.85 \times 10^{-10} \mathrm{C}$
(c) $7.85 \times 10^{-10} \mathrm{C}$
(d) $6.85 \times 10^{-10} \mathrm{C}$
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Answer:
Correct Answer: 144.(b)
Solution:
Formula:
- If $Q=$ charge on each plate, then
$ Q=C V=\frac{\varepsilon_{0} A}{d} \cdot E d=\varepsilon_{0} A E $
Here, $A=1 \mathrm{~m}^{2}, E=100 \mathrm{~N} / \mathrm{C}$
and
$ \varepsilon_{0}=8.85 \times 10^{-12} \frac{\mathrm{C}^{2}}{\mathrm{~N}-\mathrm{m}^{2}} $
So, by substituting given values, we get $Q=8.85 \times 10^{-12} \times 1 \times 100=8.85 \times 10^{-10} \mathrm{C}$
BETA
