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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. - Physics

Sum

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,

\[Q =  Q_0  .  e^{- \frac{t}{RC}} \]

\[ Q_0  =   CV   =   20 \times 6 \times  {10}^{- 6}   C\]

\[ \Rightarrow Q   = 120 \times  {10}^{- 6}  .    e^{- \frac{2 \times {10}^{- 3}}{{10}^2 . 20 \times {10}^{- 6}}} \]

\[ \Rightarrow Q = 120 \times  {10}^{- 6}  .    e^{- 1}  = 76  \mu C\]

Concept: Energy Stored in a Capacitor
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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 10 Electric Current in Conductors
Q 64 | Page 202
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