# Energy stored in Capacitor Network

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1. Oct 1, 2016

### AlisonL

1. The problem statement, all variables and given/known data
A potential difference Vab = 46.0 V is applied across the capacitor network of the following figure.

If C1=C2=4.00μF and C4=8.00μF, what must the capacitance C3 be if the network is to store 2.80×10−3 J of electrical energy?

2. Relevant equations
U=(1/2)CV^2
(1/C) = (1/c1) + (1/c2) + ... capacitors in series
C = C1 + C2 + ... capacitors in parallel

3. The attempt at a solution
U = (1/2)CV^2
2.80×10−3 = (1/2)C(46^2)
C = 2.6 μF

Capacitance of C1 and C2 = 1/(1/4 + 1/4) = 2 μF
Total capacitance:
1/2.6 = 1/(2+C3) + 1/8
C3 = 1.85 μF

I can't figure out why this answer is incorrect. Help much appreciated!

2. Oct 1, 2016

### phyzguy

I think your method is correct. Perhaps it's just round off error.? If I keep more digits, I get C = 2.6465 μF, which gives C3 = 1.955 μF. Is that the problem?

3. Oct 2, 2016

### AlisonL

That was it! Thank you so much, I didn't even consider that!