Storage of capacitors in series

AI Thread Summary
When two identical capacitors are connected in series to a 100V battery, the equivalent capacitance is halved, resulting in Ceq = C/2. The energy stored in one capacitor is U = 1/2 CV^2, and for the series combination, the energy becomes U2 = 1/2 * (C/2) * (100V)^2. This leads to the conclusion that the total energy stored in the two capacitors in series is U2 = U/2. The correct answer to the problem is therefore D) U/2. Understanding that the capacitors are identical is crucial for solving the problem correctly.
LeakyFrog
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Homework Statement


Two identical capacitors that have been discharged are connected in series across the terminals of a 100V battery. When only on of the capacitors is connected across the terminals of the battery, the energy stored is U. what is the total energy stored in the two capacitors when the series combination is connected to the battery? a) 4U, B)2U, C)U D)U/2 E)U/4

Homework Equations


U = 1/2 CV2
1/Ceq = 1/C1+1/C2

The Attempt at a Solution


The only attempt I can think to make is to get U with one capacitor in terms of C1 and then get U2 in terms of C1 and C2 and then find the ratio of them. When I do this though I keep getting U/U2 = C2/(C1+C2) Which is not any of the answers. So I'm not really sure where to go from here.
 
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U/U2 = C2/(C1+C2)
In the problem, it is given that C1 = C2.
Ceq = C/2
U1 = 1/2CV^2
U2 1/2*C/2*V^2
So U1/U2 = ...?
 


How did you know that C1 = C2? With that bit of knowledge I got D) (U/2) as the answer. Thanks for your help.
 


It says two identical capacitors.
 
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