AIIMS 2015
- Exam
- AIIMS
- Year
- 2015
- Subject
- Physics
- Chapter
- Capacitance
AIIMS 2015 · Physics · Capacitance
A 5 μ F capacitor is charged by being connected to a 3 ~V battery....
A capacitor is charged by being connected to a battery. The battery is then disconnected. If the resistance of the dielectric material between the capacitor plates is , what is the charge remaining on the capacitor after it has been disconnected ?
Options
- A
- B
- C
- D
Correct answer
B.
Step-by-step solution
\begin{aligned}& \begin{aligned}T=R C & =\left(10^9 \Omega\right)\left(5 \times10^{-6} \mathrm{~F}\right)\\
& =5 \times10^3 \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}}=\mathrm{~h}
\end{aligned}\begin{aligned}& Q_0=C V=\left(5 \times10^{-6} \mathrm{~F}\right)(3 \mathrm{~V})\\
&= 1.5 \times10^{-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 \times10^{-5} \mathrm{C}\right)\left(e^{-0.71}\right) \\
& =\left(1.5 \times10^{-5} \mathrm{C}\right)(0.49) \\
& =7.4 \times10^{-6} \mathrm{C}
\end{aligned}\end{aligned}