AP EAMCET · PHYSICS · Capacitance
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 \(0.16 \times 10^{-8} \mathrm{~F}\)
- B \(1.6 \times 10^{-8} \mathrm{~F}\)
- C \(16 \times 10^{-8} \mathrm{~F}\)
- D \(0.8 \times 10^{-8} \mathrm{~F}\)
Answer & Solution
Correct Answer
(B) \(1.6 \times 10^{-8} \mathrm{~F}\)
Step-by-step Solution
Detailed explanation
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, \(C=\frac{Q}{V}\)…
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