Energy stored in Capacitor Network

[AlisonL]

Homework Statement

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?

Homework Equations

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

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!

 

[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?

 

[AlisonL]

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