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.

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
Is there an error in this question or solution?

APPEARS IN

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