# Does charge of capacitor at constant voltage change after dielectric material is inserted?

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?

• It's definitely a much better post now. Deleting old comments... Looks like you already answered your own question too :) – DanielSank Jul 28 '16 at 6:41
• You are asked about the electric field where? – Rob Jeffries Jul 28 '16 at 21:26
• There's only gonna be between the plates, so there i suppose. – CrossNox Jul 28 '16 at 21:48
• See below, there are two values for the electric field, one in the dielectric, one in the vacuum. – Rob Jeffries Jul 28 '16 at 22:04

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...

You can think of the capacitor (when the dielectric is inserted) as being 2 separate capacitors in series: one of width $d$ and the other of width $(L-d)$, where $d$ is the width of the dielectric material and $L$ is the separation of the two metal plates.

If you have two capacitors, 1 and 2 in series, the effective capacitance is

$$C_{eff}=\frac{C_1C_2}{C_1+C_2}$$

where $C_i=\frac{\epsilon_i A}{d_i}$, $\epsilon_i$ is the permittivity of the material between the capacitor plates, $A$ is the area of the plates and $d_i$ is the separation of the plates.

You can then use this expression and the formula $Q_{new}=C_{eff}V$ to find out what the new charge on the plates will be when you have a dielectric inside. You are right that $V$ will not change, as long as you are not running out of battery.