The potential difference between the plates of a leaky (meaning that charge leaks from one plate to the other)2.0$\mathit{\mu }\mathit{F}$capacitor drops to one-fourth its initial value in 2.0 s. What is the equivalent resistance between the capacitor plates?

1. Capacitance of the given capacitor $2.0\mu F$
2. Potential difference between the plates drops to$1/4^{th}$ of the initial value within time t = 2s .

Due to the leak, the drop in the initial potential of the capacitor can also be given as the discharging condition of the capacitor. Thus, using the charge value stored within the capacitor plates, and discharging equation, the equivalent resistance of the circuit can be calculated.

Formulae:

The charge equation of a RC circuit, $\mathit{q}\mathbf{=}{\mathit{q}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{R}\mathbf{C}}$ (i)

The charge stored within the plates of a capacitor, q = CV (ii)

Substituting equation (ii) in equation (i) for the given capacitor and using the given data, the equivalent resistance between the capacitor plates can be calculated as follows:

$$\begin{gathered} V=V_0{e}^{-t/RC} \\ R=\frac{t}{C\ln \left(V_0/V\right)} \\ =\frac{2\text{s}}{\left(2\times {10}^{-6}\text{F}\right)\ln \left(\frac{V_0}{V_0/4}\right)}\left(\because V=V_0/4\right) \\ =7.2\times {10}^5\Omega \end{gathered}$$

Hence, the value of the equivalent resistance is $7.2\times {10}^5\Omega$.