A parallel-plate capacitor is filled with a dielectric material having resistivity ρ and dielectric constant K. The capacitor is charged and disconnected from the charging source. The capacitor is slowly discharged through the dielectric. Show that the time constant.

Question

Aditi Chauhan · Answer

Sol. Equation of discharging capacitor = q base 0e^-t/RC = K ϵ base 0 AV/d e^-1/(pdKϵ base 0A)/Ad = K ϵ base 0 AV/d e^-t/pKϵ base 0 ∴ τ = pK ϵ base 0 ∴ Time constant is pK ϵ base 0 is independent of plate area or separation between the plate.

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A parallel-plate capacitor is filled with a dielectric material having resistivity ρ and dielectric constant K. The capacitor is charged and disconnected from the charging source. The capacitor is slowly discharged through the dielectric. Show that the time constant.

A parallel-plate capacitor is filled with a dielectric material having resistivity ρ and dielectric constant K. The capacitor is charged and disconnected from the charging source. The capacitor is slowly discharged through the dielectric. Show that the time constant.

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A parallel-plate capacitor is filled with a dielectric material having resistivity ρ and dielectric constant K. The capacitor is charged and disconnected from the charging source. The capacitor is slowly discharged through the dielectric. Show that the time constant.

A parallel-plate capacitor is filled with a dielectric material having resistivity ρ and dielectric constant K. The capacitor is charged and disconnected from the charging source. The capacitor is slowly discharged through the dielectric. Show that the time constant.

1 Answers

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Other Related Questions on Electromagnetic Induction

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