Question:
A parallel plate capacitor with plates of area 1 m2 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 \(\varepsilon_0=8.85\times10^{-12}\ \frac{\mathrm{C^2}}{\mathrm{N\,m^2}}\))
A parallel plate capacitor with plates of area 1 m2 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 \(\varepsilon_0=8.85\times10^{-12}\ \frac{\mathrm{C^2}}{\mathrm{N\,m^2}}\))
(Take \(\varepsilon_0=8.85\times10^{-12}\ \frac{\mathrm{C^2}}{\mathrm{N\,m^2}}\))
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Field between the plates of a capacitor is \(E=\frac{\sigma}{\varepsilon_0}=\frac{Q}{\varepsilon_0A}\), so \(Q=\varepsilon_0EA\).
Updated On: Nov 5, 2026
- \(9.85\times10^{-10}\) C
- \(8.85\times10^{-10}\) C
- \(6.85\times10^{-10}\) C
- \(7.85\times10^{-10}\) C
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The Correct Option is B
Solution and Explanation
\(E=\frac{\sigma}{\varepsilon_0}=\frac{Q}{\varepsilon_0A}\Rightarrow Q=\varepsilon_0EA\).
\(Q=8.85\times10^{-12}\times100\times1=8.85\times10^{-10}\) C.
Answer: (2) \(8.85\times10^{-10}\) C
\(Q=8.85\times10^{-12}\times100\times1=8.85\times10^{-10}\) C.
Answer: (2) \(8.85\times10^{-10}\) C
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