The electric field between the two parallel plates of a capacitor of capacitance $$1.5\mu\text{F}$$ drops to one third of its initial value in $$6.6\mu s$$ when the plates are connected by a thin wire. The resistance of this wire is ______ $$\Omega$$. (Given, $$\log 3 = 1.1$$)

The capacitor ($$C = 1.5\mu$$F) discharges through a wire of resistance $$R$$ and the electric field drops to $$\frac{1}{3}$$ of its initial value in $$6.6\mu$$s.

Because the electric field is proportional to the voltage across the capacitor, $$E \propto V$$, the voltage during discharge follows

$$ V = V_0 e^{-t/RC} $$

When $$V = V_0/3$$, we have

$$ \frac{1}{3} = e^{-t/RC} $$ $$ \ln 3 = \frac{t}{RC} $$ $$ RC = \frac{t}{\ln 3} $$

Substituting $$t = 6.6 \mu$$s and $$\ln 3 = \log_e 3 = 1.1$$ gives

$$ RC = \frac{6.6 \times 10^{-6}}{1.1} = 6 \times 10^{-6} \text{ s} $$ $$ R = \frac{6 \times 10^{-6}}{1.5 \times 10^{-6}} = 4 \text{ }\Omega $$

Hence, the answer is 4.