A parallel plate capacitor is charged to $ 60 \mu \mathrm{C} $. Due to a radioactive source, the plate loses charge at the rate of $ 1.8 \times 10^{-8} \mathrm{C} \mathrm{s}^{-1} $. The magnitude of displacement current is

Question

A parallel plate capacitor is charged to $ 60 \mu \mathrm{C} $. Due to a radioactive source, the plate loses charge at the rate of $ 1.8 \times 10^{-8} \mathrm{C} \mathrm{s}^{-1} $. The magnitude of displacement current is, $ 1.8 \times 10^{-8} \mathrm{C} \mathrm{s}^{-1} $, $ 3.6 \times 10^{-8} \mathrm{C} \mathrm{s}^{-1} $, $ 4.1 \times 10^{-11} \mathrm{C} \mathrm{s}^{-1} $, $ 5.7 \times 10^{-12} \mathrm{C} \mathrm{s}^{-1} $

MARKS App · Accepted Answer

Given that, d q d t = 1 . 8 × 10 - 8 the displacement current is I d = d q d t I d = 1 .8 × 10 - 8 C s - 1

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