Question

The electric field between the two parallel plates of a capacitor of 1.5 μF capacitance drops to one third of its initial value in 6.6 μs when the plates are connected by a thin wire. The resistance of this wire is .............. Ω. (Given, log 3 = 1.1)

Answer 1

Correct Answer: 4

\[ E = \frac{E_0}{3} \implies V = \frac{V_0}{3} \]

\[ \frac{V_0}{3} = V_0 e^{-\frac{t}{\tau}} \]

\[ t = \tau \ln 3 \]

\[ 6.6 \times 10^{-6} = R (1.5 \times 10^{-6})(1.1) \]\[ R = \frac{6.6}{1.5} = 4 \, \Omega \]

Answer 2

Step 1: Model the discharge
When the plates are shorted by a wire of resistance \(R\), the capacitor (capacitance \(C=1.5\,\mu\text{F}\)) discharges through \(R\) with time constant \(\tau=RC\). The voltage (and hence the electric field between the plates) decays exponentially: \[ E(t)=E_0\,e^{-t/(RC)}. \] Given that the electric field drops to one third in \(t=6.6\,\mu\text{s}\): \[ \frac{E(t)}{E_0}=\frac{1}{3}=e^{-t/(RC)} \quad\Longrightarrow\quad \frac{t}{RC}=\ln 3. \]

Step 2: Solve for the resistance
\[ R=\frac{t}{C\,\ln 3}. \] Insert the values \(t=6.6\times 10^{-6}\,\text{s}\), \(C=1.5\times 10^{-6}\,\text{F}\), and (as given) \(\ln 3\approx 1.1\): \[ R=\frac{6.6\times 10^{-6}}{(1.5\times 10^{-6})\times 1.1} =\frac{6.6}{1.65} =4\,\Omega. \]

Step 3: Note on the logarithm
The problem’s “log 3 = 1.1” is intended as the natural logarithm \(\ln 3\approx 1.1\) (not base-10). Using base-10 would give an incorrect value.

Final answer

4