I have a planar geometry capacitor, connected to a battery that supplies V volts. Initially there is vacuum/air in between both plates.
Afterwards, some dielectric material is inserted in between both plates, filling the capcitor as follows, with the green thing being the dielectric :
I am asked to compare the initial and final electric fields. The plates are never disconnected from the battery.
My question is the following: Should the charge of each plate be the same in the initial and final states?
My thoughts on it: Since the plates are never disconnected, the voltage will remain constant, and so Qf/Cf=Qo/Co. Since i put some dielectric, as far as i understand, Cf should be higher than Co, and so Qo should be proportionally higher than Qf. Also, based on Spirko's answer here, the battery will change the charge in the plates to mantain the voltage.
Is this correct?
Best Answer
Yes, you are correct. The potential difference between the plates remains the same, so the charge is increased by the same factor as the capacitance is increased - let's call this factor $k$ and let the dielectric have relative permittivity $\epsilon$ and fill a fraction $f$ of the plate separation $d$.
So $V = Q_0 d/\epsilon_0 A$, where $A$ is the plate area and the electric field initially between the plates $E_0 = Q_0/\epsilon_0 A$, which you can confirm with Gauss's law.
Then insert the dielectric. $V$ stays the same, but the E-field now has two values, one in the vacuum $E_v$ and a lower value inside the dielectric $E_d$. The displacement field is the same throughout the space between the plates and is given by Gauss's law as $D=Q_f/A$.
From the relation between E-field and potential $$V = (1-f)d E_v + fdE_d$$ and $$E_v =\epsilon E_d = D/\epsilon_0 = Q_f/\epsilon_0 A.$$
Putting this all together: $$ E_0 d = (1-f)d \epsilon E_d +fd E_d$$ $$ E_d = \frac{E_0}{(1-f)\epsilon +f}$$
All other quantities of interest follow from this...