Physics · 2019
JEE Main · 12 January 2019, Shift 2 · Q14
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…
A parallel plate capacitor with plates of area $\displaystyle 1 \mathrm{~m}^2$ each, are at a separation of $\displaystyle 0.1$ m . If the electric field between the plates is $\displaystyle 100$ N/C, the magnitude of charge on each plate is :
$$\left(\text { Take } \epsilon_0=8.85 \times 10^{-12} \frac{\mathrm{C}^2}{\mathrm{~N}-\mathrm{m}^2}\right)
$$
Official answer
(3)
$\displaystyle 8.85 \times 10^{-10} \mathrm{C}$
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.