# What happens to a system of capacitors when a dielectric is inserted into one of them?

sorry for my English. While doing some homework I've stumbled upon this "circuit":

the system is connected to the same potential difference through a generator that is connected to points A and B.

At first the problem asks to calculate the total capacitance of the system before and after the insertion of the dielectric in C3; when asked to calculate the potential difference for C3 only, the solution:

1. Calculates the global charge because we know both the potential difference and the total capacitance of the system;
2. Equals the total charge and the charge of the C3 capacitor: $$Q_{tot} = Q_{3}$$.
3. Since we know everything else, we can easily find the potential in C3.

But why does this happen? The solution gives only this as "explanation":

"The final charge on the capacitor 3 will be equal to the total charge of the system."

I have already tried to wrap my head around this and I can't get it; maybe it has something to do with the fact that the first two capacitors are in parallel? Maybe they lose all their charge? But why does this happen?

Thank you

Consider the part of the circuit I have outlined in red:

This part of the circuit is isolated from A and B so no charge can flow off or onto it. That means if its total charge is zero before we apply any potential its total charge must remain zero after we apply a potential. All the potential does is cause a charge separation. Fr example if B is the positive electrode then the charge would look like this:

The charge on both $$C_1$$ and $$C_2$$ combined adds up to $$+Q$$. How much of this is on each capacitor depends on the relative values of their capacitances. Anyhow, since the charges on the plates of a capacitor are equal and opposite we can fill in the rest of the charges:

Hence the claim that the charge $$Q$$ on $$C_3$$ is equal to the charge on $$C_1$$ and $$C_2$$ combined, and equal to the charge when the whole system is combined into a single capacitor.

Simple, the charge measured in coulombs is the current in amperes times the time in seconds, so Q=I*t. since C3 is connected in series with both C1 and C2 this means that any current flowing through C1 and C2 will also flow through C3.

Consider this example; lets say instead of a voltage potential between A and B we had a constant current source with an output of 1A, now we turn the current source on at t=0 and we turn it off again after 10s. What is the current through C3 in the 10s where we have the current source switched on?? well all the current going into A goes through C3 and hence the current through C3 is 1A for the full 10s that we have the current source switched on. Now after 10s what is the charge that has build op on C3?? well it is simply 1A*10s=10C. Now lets instead consider all three capacitors C1, C2 and C3 as just one capacitance, and see what happens when we switch our 1A current source on for 10s; all the current flowing into A flows through the combined capacitor C1,C2,C3 and hence the charge that builds up on the combined capacitor is still 1A*10s=10C..

In other words if capacitors are coupled in series the current through them is always the same according to KCL and hence the charge that builds up on them is the same.

another true argument about the circuit in question would be that the charge on C1+C2 is also equal to the charge on C3, and is also equal to the total charge in the circuit.

Note that saying that the charge on C3 is equal to the charge on C1+C2 and is equal to the total charge on the circuit is not the same as saying that all the energy is contained in C3, there will be a voltage both across C3 and across C1+C2 and the total voltage between A and B will be larger than the voltage across C3 and hence the total energy in the combined capacitor will be the energy in C3 + energy in C1 + energy in C2. and the energy is equal to 1/2*V*Q.