The electric field in space between the plates of a parallel plate capacitor (each of area 2.5 10^ -3 ~m ^ 2 ) is changing at the rate of 4 10^ 6 Vm ^ -1 ~s ^ -1 . The displacement current between the plates of the capacitor is:

The electric field in space between the plates of a parallel plate capacitor (each of…

Physics · Class 12 · CBSE — Electromagnetic Waves

The electric field in space between the plates of a parallel plate capacitor (each of area \(2.5 \times 10^{-3} {~m}^{2}\) ) is changing at the rate of \(4 \times 10^{6} {Vm}^{-1} {~s}^{-1}\) . The displacement current between the plates of the capacitor is:

\(1.8 \times 10^{-5} {~A}\)
\(3.47 \times 10^{-6} {~A}\)
\(8.85 \times 10^{-8} {~A}\)
\(6.32 \times 10^{-4} {~A}\)

Answer

(C) 8.85 10^ -8 ~A

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The electric field in space between the plates of a parallel plate capacitor (each of…

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