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Question
Consider the situation shown in the figure. The switch S is open for a long time and then closed. (a) Find the charge flown through the battery when the switch S is closed. (b) Find the work done by the battery.(c) Find the change in energy stored in the capacitors.(d) Find the heat developed in the system.
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Solution
Since the switch is opened for a long time, the capacitor is in series.
`therefore C_(eq) = C/2`
When the switch is closed, the charge flown from the battery is given by
`q = C/2 xx ε =( Cε)/2`
(b) Work done, `W = q xx V`
`⇒ q = C_ε/2 xx ε = (Cε^2)/2`
(c) The change in the energy stored in the capacitors is given by
`ΔE = 1/2 C_(eq) xx V_2`
`= 1/2 xx C/2 xx ε^2 = 1/4 Cε^2`
(d) The heat developed in the system is given by
H = ΔE
`⇒ H = 1/4 Cε^2`
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