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I've never done exercises like that and I don't have the results.

## Homework Statement

A generator which mantains a constant potential difference of 400V is connected to two capacitors in series C1 = 500pF and C2=1000pF. A dielectric is inserted into C1 (##\epsilon_r = 4##).

Calculate the charge variation on C2, the polarization charge on both faces of the dielectric and the energy supplied by the generator.

## The Attempt at a Solution

I calculated the ##C_eq## of the system of the capacitors without the dielectric.

## \frac{1}{C_eq} = \frac{1}{C1} + \frac{1}{C2} ##

## C_eq = 333 pF ##

Since the capacitors are connected in series, q is the same on both capacitors.

##q = CV = 1.33x10^{-7}C ##

Now I calculate C1 with the dielectric.

## C1 = \epsilon_r C1 = 2x10^-9 F ##

The new ##C_eq## is ##6.66x10^-10F##.

##q = CV = 2.66x10^{-7} C##

I can calculate easily the variation of the charge.

The potential difference is constant, the capacity (as I expected) increased. q could only increase.

Now I find the polarization charge on both faces.

I find it with the vector ##\vec{P} = \epsilon_0 \chi \vec{E}##

## P = \sigma_p = \frac{q_p}{S} = \epsilon_0 (\epsilon_r -1) \frac{q}{S} ##

##q_p## is 1.99x10^-7. It is +qp near the negative face of the capacitor, and it is -qp near the positive one.

The energy is

##U = \frac{1}{2} CV^{2} ##

Could you please tell me if that's right?

If not, please tell me something to think about!

Thank you