# Voltage in capacitors

1. Two capacitors, C1 = 60.0 nF and C2, were connected in parallel with a power supply with E = 12.0 V. The two capacitors were then disconnected from the power supply and from each other and were reconnected with the positive terminal of one connected to the negative terminal of the other capacitor. After reconnection, the voltage across C2 was 4.00 V. C2 is less than C1.
Calculate C2

3. well originally it says they are in parallel
so

q1 = 60*12
q2 = C2*12

After reconnection
q1+q2 = q1' + q2'

q1' = q2' ? because they are in series? ... if i use this i get the wrong answer
but if i use this

q1+q2 = q1' + q2'
60*12 + c2*12 = 60*8 + c2*4

8 because still basically the remaining voltage should be across the other capictor (since im thinking they are in series)

I solve for C2 = and i get the NEGATIVE RIGTH answer -30nF..while answer is +30nF

Help !!!!!!!!

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alphysicist
Homework Helper
After the capacitors are reconnected to each other, they will be in parallel, and so the potential difference across each will be the same (4V).

Also, when using the charge formula

q1 + q2 = q1' + q2'

if q1 is positive, then q2 will be negative (because you're connecting the positive plate of one capacitor to the negative plate of the other, and vice versa). By connecting the plates together like this, there will be some cancellation of the charges, and after connection q1' and q2' will both be positive (since q1 is positive and larger in magnitude than q2, the 'leftover' charge after cancellation that is shared between the plates will be positive).