When 1.0 × 10^12 electrons are transferred from one conductor to another, a potential difference of 10 V appears between the conductors. Calculate the capacitance of the two-conductor system.
Hrishant Goswami , 10 Years ago
Grade 10
1 Answers
Deepak Patra
Last Activity: 10 Years ago
Sol. Given that
Number of electron = 1 × 10^12
Net charge Q = 1 × 10^12 × 1.6 × 10^–19 = 1.6 × 10^–7 C
∴ The net potential difference = 10 L.
∴ Capacitance – C = q/v = 1.6 * 10^-7/10 = 1.6 * 10^-8 F.
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