# Two capacitors sharing charge

My two capacitors:

I have these two capacitors ($C_1 = 3\mu F, C_2 = 4\mu F$) both initially under 19 volts. Then, I added a dielectric with $k=4$ at $C_1$ and entire air gap got filled. How can I calculate the potential between the two capacitors, or the electric field of each one of them or the charge of each one of them? The distance between the two of plates of both capacitors is $0.003\ m$.

Why the new voltage is gonna be $q_{total}/(4\mu\times4+3\mu)$ and not $q_{total}/(4\mu+3\mu\times4)$? I'm inserting the dielectric in the $C_1$ not in $C_2$...

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I'm pretty sure the second one is right. Are you sure you're reading the answer correctly? –  Ataraxia Mar 23 '13 at 23:37
I'm sure about that. I tested the two answers. Only the first one answer is accepted. It seems that is a different situation when I try to mount a equivalent circuit with C1 = 12u with 19/4V previously charged and C2 = 4u with 19V previously charged. In this circuit I get the first one result. –  João Paulo Oliveira Fernandes Mar 24 '13 at 0:44

This is a system of two capacitors in parallel, so the total charge is given by $$Q_\mathrm{tot} = (C_1 + C_2)V ~.$$ When a dielectric is inserted into capacitor 1, it's capacitance changes: $C_1 \to \kappa\,C_1$. However, the total charge is conserved. Thus $$Q_\mathrm{tot} = (\kappa\,C_1 + C_2)V' ~.$$ This implies that \begin{align} (C_1 + C_2)V &= (\kappa\,C_1 + C_2)V' \\ \Rightarrow V' &= \frac{(C_1 + C_2)V}{\kappa\,C_1 + C_2} \\ \therefore V' &= \frac{Q_\mathrm{tot}}{\kappa\,C_1 + C_2} ~. \end{align} If this result is not working wherever you are submitting it, then it may be that the website is wrong, or you are misreading the question.