A 6.0-μF capacitor charged to 50 V and a 4.0-μF capacitor charged to 34 V are connected to each other, with the two positive plates connected and the two negative plates connected. What is the total energy stored in the 6.0-μF capacitor at equilibrium?
A) 6.1 mJ
B) 5.7 mJ
C) 6.6 mJ
D) 7.0 mJ
E) 3.8 mJ

Question

Quizplus · Accepted Answer

To find the equilibrium state, we can equate the total energy stored in the capacitors with the initial energy they had when they were not connected. The initial energy stored in the 6.0-μF capacitor is: 
$E_1 = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} (6.0	imes 10^{-6})(50)^2 = 7.5	imes 10^{-3}$ J
The initial energy stored in the 4.0-μF capacitor is:
$E_2 = \frac{1}{2} C_2 V_2^2 = \frac{1}{2} (4.0	imes 10^{-6})(34)^2 = 2.3	imes 10^{-3}$ J
When the two capacitors are connected, charge will flow between them until their potentials are the same. Let the final potential be V. Then, the final charge on the 6.0-μF capacitor is Q = CV = (6.0×10−6)×V, and the final charge on the 4.0-μF capacitor is Q = CV = (4.0×10−6)×V. So, the total charge transferred between the capacitors is:
$\Delta Q = Q_1 - Q_2 = (6.0	imes 10^{-6})V - (4.0	imes 10^{-6})V = 2.0	imes 10^{-6}V$
Applying conservation of charge, this must be equal to the charge on the final combination of capacitors:
$Q_T = C_T V$
where $C_T$ is the equivalent capacitance of the combination, given by:
$\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2}$
Substituting the values gives $C_T = (6.0	imes 10^{-6})(4.0	imes 10^{-6})/(10.0	imes 10^{-6}) = 2.4	imes 10^{-6}$ F.
Thus, we have $2.0	imes 10^{-6}V = (2.4	imes 10^{-6})V^2/2$, and solving for V gives $V = \frac{5}{3}$ V.
The energy stored in the 6.0-μF capacitor at equilibrium is:
$E = \frac{1}{2} C_1 V^2 = \frac{1}{2} (6.0	imes 10^{-6})\left(\frac{5}{3}ight)^2 = 5.7	imes 10^{-3}$ J = 5.7 mJ (to 2 significant figures)

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