Resistance of wire in discharging capacitor – Rankers Physics

Charging and Discharging of Capacitors: Practice Problem & Solution

The electric field between the plates of a parallel-plate capacitor of capacitance \(2.0~\mu\text{F}\) drops to one third of its initial value in \(4.4~\mu\text{s}\) when the plates are connected by a thin wire. Find the resistance of the wire.
\(0.5~\Omega\)
\(0.1~\Omega\)
\(2~\Omega\)
\(1~\Omega\)

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Charging and Discharging of Capacitors. Understanding the underlying formula is key to arriving at the correct answer below:

The electric field in a discharging capacitor drops as \(E = E_0 e^{-t/RC}\). Given \(E = E_0/3\), we have \(RC = \frac{t}{\ln 3}\). Solving for \(R = \frac{4.4 \times 10^{-6}}{2.0 \times 10^{-6} \times 1.1} = 2~\Omega\).

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