Capacitors in parallel and dieletric [closed]

I'm trying to solve this problem:

Two capacitors of capacitance $C_1=200pF$ and $C_2=1000pF$ are connected in parallel and loaded to a potential difference of $400V$. Subsequently the space between the plates of $C_1$ is completely filled with distilled water (with $ε = 80$). Calculate the variation of the difference $ΔV$ voltage across $C_2$, the polarization charge $q_p$ on the faces of dielectric, the variation of electrostatic energy of the system $ΔU$.

Now in order to obtain a lot of information I calculated $Q_1=8.0 \times 10^{-8}C$, $Q_2=4.0\times 10^{-7}C$, $C_{total}=1.2\times 10^{-9}F$ and $Q_{total}=4.8\times 10^{-7}C$, then I calculated the new capacitance of $C_1$ because of the dieletric so $C_{1ε}=1.6\times 10^{-8}$ Here I'm stuck, how can I calculate the variation of the difference of potential $ΔV$ voltage across $C_2$ since the capacitors are in parallel? Then how can I calculate the polarization charge $q_p$ on the faces of dielectric and the variation of electrostatic energy of the system $ΔU$. I also want to understand how the system change after the introduction of the dieletric. Can someone guide me to the solution in a fine way?

closed as off-topic by John Rennie, Dan, Brandon Enright, Kyle Kanos, Emilio PisantyJan 12 '14 at 23:26

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Well first of all the major points here are potential difference across both capacitors will remain same after insertion of dielectric in $C1$, and the total charge of the system will remain constant.

Now in $C1$ after insertion of dielectric, calculate change of capacitance, assume that it has new potential $V1$ and charge $Q1$, assume potential and charge for 2nd capacitor too as $V2$ and $Q2$.

Equate $V1$ and $V2$, also equate $Q1 + Q2$ with initial total charge of system. This way you can calculate $V$ across capacitors and $Q1$ and $Q2$.

Now to calculate polarisation charge, check the electric field $E1 (without dielectric)$ Now find electric field $E2 (with dielectric)$ this difference has come because of insertion of dielectric so the electric field of dielectric (calculate for a hypothetical parallel plate capacitor) is equal to the difference in electric fields, this will give you tue induced charge.

Lastly, since you know initial and final potential and charges, apply formula for potential energy of capacitors both before and after insertion of dielectric and take the difference, this is the change in potential energy.

The system undergoes changes as the following, once the capacitors are charged, when dielectric is being inserted in the capacitor $C1$ charges are induced in it and it is pulled in by electrostatic interactions, this process consumes some energy and this changes the potential. Also since the capacitors are in parallel and maintain same potential, the charges travel from one capacitor to another after insertion of dielectric to eliminate the potential difference between capacitors.

• Thank you very much! But I've not been able to find the polarisation charge, how can I calculate the electric field? I don't have enough infos since $E=σ/\epsilon_0$ and $σ=Q/S$ where $S$ is the area of the plates of capacitors... – Dipok Jan 12 '14 at 17:00
• Just replace the difference in electric fields (initial and final) in place of $E$ in your equation this should give you the field in dielectric and hence the induced charge on its surfaces. – Rijul Gupta Jan 12 '14 at 17:02