# Homework Help: The potential difference in a capacitor with dielectric

1. Jan 27, 2010

### Ripperbat

1. The problem statement, all variables and given/known data

Two identical capacitors are connected in parallel, charged by a battery ov Voltage V, and then isolated from the battery. One of the capacitors is then filled with a material of relative permittivety E.

2. Relevant equations

What is the final potential difference across it's plates?

3. The attempt at a solution

Potential difference?

Q = Charge

2C in parallel

Q1 = C1V
Q2 = C2V

C1=C2=C

Q=Q1+Q2 = CV+CV=2CV (on each capacitor)

Q=2C0V -> C0 =( Q / (2V) )

The battery is removed, one capacitor is filled with dielectric with relative permittivety E.

C = EC0

Q = CfVf = CiVi

Vf = (Ci/Cf)*Vi = (CiVi)/Cf = (2C0V)/EC0 = 2V/E

The correct solution should look like: (2V)/(E+1)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 27, 2010

### ehild

The sum of the charge on both capacitors stays the same as before: Q=2VCo, but the new resultant capacitance is (E+1)Co, so the new voltage is V'= Q/Cresultant= 2VCo/((E+1)Co)

ehild

3. Jan 27, 2010

### Ripperbat

So what they mean by isolating the circuit from the battery is that it still remains a closed circuit, hence we have to take both capacitors into account when solving this?

Last edited: Jan 27, 2010