Solution:
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 \(R = \frac{4.4 \times 10^{-6}}{2.0 \times 10^{-6} \times 1.1} = 2~\Omega\).
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.
Leave a Reply