Example

Question:

(a) A 900 pF capacitor is charged by a 100 V battery [Fig. 2.31(a)]. How much electrostatic energy is stored by the capacitor?
(b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor [Fig. 2.31(b)]. What is the electrostatic energy stored by the system?

Solution:

(a) The charge on the capacitor is \[ Q = CV = 900 \times 10^{-12}\,\text{F} \times 100\,\text{V} = 9 \times 10^{-8}\,\text{C} \] The energy stored by the capacitor is \[ U = \frac{1}{2} CV^2 = \frac{1}{2} Q V = \frac{1}{2} \times 9 \times 10^{-8}\,\text{C} \times 100\,\text{V} = 4.5 \times 10^{-6}\,\text{J} \] (b) In the steady state, both capacitors have the same potential difference. By conservation, \[ Q' = Q/2, \qquad V' = V/2 \] The total energy is \[ U_{\text{system}} = 2 \times \frac{1}{2} Q' V' = \frac{1}{4} Q V = 2.25 \times 10^{-6}\,\text{J} \] Thus the final energy is only half the initial. The missing energy is lost as heat and electromagnetic radiation during the transient period when current flows from the first to the second capacitor.