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Question
A 20 μF capacitor is joined to a battery of emf 6.0 V through a resistance of 100 Ω. Find the charge on the capacitor 2.0 ms after the connections are made.
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Solution
The growth of charge across a capacitor,
`Q=CV(1-e^-t/(RC))`
\[ Q_0 = CV = 20 \times 6 \times {10}^{- 6} C\]
= `20 xx 6 xx 10 ^(-6) (1 - e (- 2 xx (10)^-3)/((10)^2 . 20 xx (10)^-6))`
= `12 xx 10 ^ -5 (1 - e ^ -1)`
= `7.12 xx 0.63 xx 10 ^-5`
= `7.56 xx 10 ^-5`
= `75.6 xx 10 ^-6`
= 76 μc.
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