Electrostatics

JEE Main level, single correct, medium

119 of 311

Question 119 of 311

A parallel plate capacitor with plates of area 1m21{m}^{2} each, are at a separation of 0.1m.0.1m. If the electric field between the plates is 100N/C,100 N/C, the magnitude of charge on each plate is: (Take∈0=8.85×10−12C2N-m2)(\text{Take} {\in }_{0}=8.85\times {10}^{-12}\dfrac{{C}^{2}}{N\text{-}{m}^{2}})

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Answer: (A) 8.85x10−10C8.85x{10}^{-10}C

E=σϵ0=QAϵ0=100V/mE=\dfrac{\sigma }{{\epsilon }_{0}}=\dfrac{Q}{A{\epsilon }_{0}}=100V/m

Q=100×Aϵ0Q=100\times A{\epsilon }_{0}

⇒Q=100×1×8.85×10−12C\Rightarrow Q=100\times 1\times 8.85\times {10}^{-12}C

=8.85×10−10C=8.85\times {10}^{-10}C

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