How Do You Calculate the Capacitance of C2 After Reconnection?

[salman213]

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 I am thinking they are in series)

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

Help !

 

[alphysicist]

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).

 

[Moetasim]

The correct way to solve this problem is by using the equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. Since the capacitors are in parallel, the total charge on both capacitors will be equal to the charge on C1 and the charge on C2.

Initially, when the capacitors are connected to the power supply, the total charge on both capacitors is given by:

Q = C1*E + C2*E

where E is the voltage of the power supply.

After disconnection and reconnection, the charge on both capacitors remains the same. However, the voltage across C1 and C2 changes. Since they are now connected in series, the total voltage across both capacitors will be equal to the sum of the individual voltages:

V = V1 + V2

We know that V1 = E (since it is still connected to the power supply) and V2 = 4.00 V (given in the problem). Therefore, we can write:

V = E + 4.00 V

Substituting this into the equation Q = CV, we get:

C1*E + C2*E = C1*V + C2*V

Solving for C2, we get:

C2 = (C1*E - C1*V)/(V-E)

Substituting the given values, we get:

C2 = (60*12 - 60*8)/(8-12) = 30 nF

Therefore, the value of C2 is +30 nF, which is the correct answer.