Q112 / 180 · AIIMS 2015 · Physics
A $5 \mu \mathrm{F}$ capacitor is charged by being connected to a $3 \mathrm{~V}$ battery. The battery is then disconnected. If the resistance of the dielectric material between the capacitor plates is $10^9 \Omega$, what is the charge remaining on the capacitor $1 \mathrm{~h}$ after it has been disconnected ?
- A$1.5 \times 10^{-5} \mathrm{C}$
- Correct answerB$7.4 \times 10^{-6} \mathrm{C}$
- C$5.4 \times 10^{-6} \mathrm{C}$
- D$8.4 \times 10^{-5} \mathrm{C}$
Solution
$$
\begin{aligned}
& \text { } \begin{aligned}
T=R C & =\left(10^9 \Omega\right)\left(5 \times 10^{-6} \mathrm{~F}\right) \\
& =5 \times 10^3 \mathrm{~s}
\end{aligned} \\
& =\frac{5 \times 10^3 \mathrm{~s}}{(60 \mathrm{~s} / \mathrm{min})(60 \mathrm{~min} / \mathrm{h})}=\frac{5000 \mathrm{~s}}{3600 \mathrm{~s} / \mathrm{h}}=1.4 \mathrm{~h}
\end{aligned}
$$
The initial charge on the capacitor is
$$
\begin{aligned}
& Q_0=C V=\left(5 \times 10^{-6} \mathrm{~F}\right)(3 \mathrm{~V}) \\
&= 1.5 \times 10^{-5} \mathrm{C} \\
& \text { After } t=1 \mathrm{~h}, t / T=(1 \mathrm{~h}) /(1.4 \mathrm{~h}) \\
&=0.71 \\
& \text { and } \begin{aligned}
Q & =Q_0 e^{-t / T} \\
& =\left(1.5 \times 10^{-5} \mathrm{C}\right)\left(e^{-0.71}\right) \\
& =\left(1.5 \times 10^{-5} \mathrm{C}\right)(0.49) \\
& =7.4 \times 10^{-6} \mathrm{C}
\end{aligned}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.