JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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
- A \(7.85\times 10^{-10}\,C\)
- B \(6.85\times 10^{-10}\,C\)
- C \(8.85\times 10^{-10}\,C\)
- D \(9.85\times 10^{-10}\,C\)
Answer & Solution
Correct Answer
(C) \(8.85\times 10^{-10}\,C\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}=\frac{\mathrm{q}}{\mathrm{A} \varepsilon_{0}}\) \(\Rightarrow \quad \mathrm{q}=\mathrm{E} \mathrm{A} \varepsilon_{0}\) \(\Rightarrow \quad \mathrm{q}=100 \times 1 \times 8.85 \times 10^{-12}\) \(\Rightarrow \mathrm{q}=8.85 \times 10^{-10}\, \mathrm{C}\)
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