When 1.0 × 1012 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 Answer
Deepak Patra
Last Activity: 10 Years ago
Sol. Given thatrnNumber of electron = 1 × 10^12rnNet charge Q = 1 × 10^12 × 1.6 × 10^–19 = 1.6 × 10^–7 Crn∴ The net potential difference = 10 L.rn∴ Capacitance – C = q/v = 1.6 * 10^-7/10 = 1.6 * 10^-8 F.
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