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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 : $$\left(\text { Take } \epsilon_0=8.85 \times 10^{-12} \frac{\mathrm{C}^2}{\mathrm{~N}-\mathrm{m}^2}\right) $$
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.