Q 12-02-068JEE MainJEE Main 2024 (5 Apr, Shift 1)Medium
The electric field between the two parallel plates of a capacitor of $1.5\ \mu\text{F}$ capacitance drops to one third of its initial value in $6.6\ \mu\text{s}$ when the plates are connected by a thin wire. The resistance of this wire is ______ $\Omega$. (Given $\ln3 = 1.1$)
Numerical value type. Enter your answer.
Answer: 4
The field is proportional to the charge, which decays as $q = q_0e^{-t/RC}$:
$$\frac13 = e^{-t/RC} \Rightarrow t = RC\ln3$$
$$R = \frac{6.6\times10^{-6}}{1.5\times10^{-6}\times1.1} = 4\ \Omega$$
Solution by Sreeraj P, M.Sc Physics