Q146 / 160 · AP EAMCET 2022 (07 Jul Shift 1) · Physics
In a parallel plate capacitor, if $10^{12}$ electrons pass from one plate to another, a potential difference of $10 \mathrm{~V}$ is developed across the plates. The capacitance of the capacitor is
- A$0.16 \times 10^{-8} \mathrm{~F}$
- Correct answerB$1.6 \times 10^{-8} \mathrm{~F}$
- C$16 \times 10^{-8} \mathrm{~F}$
- D$0.8 \times 10^{-8} \mathrm{~F}$
Solution
Charge on each plate of capacitor has a magnitude,
$Q=N . e$
Here, $N=10^{12}, e=1.6 \times 10^{-19}$
Potential difference between plates,
$V=10 \mathrm{~V}$
Capacitance of capacitor, $C=\frac{Q}{V}$
$\Rightarrow C=\frac{N e}{V}=\frac{10^{12} \times 1.6 \times 10^{-19}}{10}=1.6 \times 10^{-8} \mathrm{~F}$
Hence, the correct answer is option: 'B'.