# Capacitors in series with dielectric: Charge distribution

If you derive the formula

$$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$$

for two capacitors in series one usually assumes that the charge on $C_1$ is the same as the charge on $C_2$.

Now if you have one capacitor partially filled with a dielectric one can think of this as two (or three if the dielectric is in the middle) capacitors in series, one with the other without dielectric. Then the formula above is applied.

However if you consider this situation the surface charge density on the dielectric surface is less than the surface charge density on the plates.

This seems to contradict the assumption made in the derivation of the formula above. How to solve this apparent contradiction?

• What matters is the free charge: this is what enters in an expression like $Q=CV$; the surface of the dielectric contains bound charges (resulting from the polarization of the material). – ZeroTheHero Mar 12 '17 at 19:58
• Ok, so in the equivalent series of capacitors, there would be indeed the same charge on the plate of the capactor with the dielectric as on the other capacitor? – Julia Mar 12 '17 at 20:09
• added an answer. – ZeroTheHero Mar 12 '17 at 21:54

I don't have exactly the figures to deal with your specific examples but these ones will be useful. Just replace the second medium with vacuum to deal with your specific example. You can also adjust the argument to have any thickness of each medium.

You are correct that, on the surface of the dielectric, the surface charge density $\sigma_i$ will be less than the surface charge density $\sigma$ on the conducting plates. If there were only one type of material the situation would be like this:

It's not hard to see how this generalizes to two materials. Indeed, the argument about surface charges does not depend on the number of materials, as you have correctly guessed.

But one does not use $\sigma_i$: instead one must know $V_1$ and $V_2$. To get these one needs the electric fields in medium $1$ and $2$, given respectively by $$E_1=\frac{\sigma}{\epsilon_1}\, ,\qquad E_2=\frac{\sigma}{\epsilon_2}$$ where $\sigma$ is still the surface charge density on the conducting plates. The effect the charges induced in the dielectrics is to reduce the fields $E_1$ and $E_2$, and this is captured by $\epsilon_1$ and $\epsilon_2$. The fields $E_1$ and $E_2$ are the net fields, resulting from the superposition of the field created by the plates and the field from the induced charges with surface density $\sigma_i$. Since the $\vec E_k$ field are constant in each material (by Gauss's law), we have \begin{align} V_1&= E_1\times \frac{d}{2}\, ,\qquad\qquad\qquad\qquad V_2=E_2\times \frac{d}{2}\, ,\\ V&=V_1+V_2=\sigma\frac{d}{2}\left(\frac{1}{\epsilon_1}+\frac{1}{\epsilon_2}\right)\, . \end{align} You can see here again that everything is expressed in terms of the charge density $\sigma$ on the conducting plates; the fact there are bound charges on the dielectrics is included in the values of the permittivities $\epsilon_1$ and $\epsilon_2$.

Since now $C=Q/V$ where $Q=\sigma A$ is the free charge on the plates of area $A$, you find $$\frac{1}{C}=\frac{V}{Q}=\frac{V_1}{\sigma A}+\frac{V_2}{\sigma A}$$ and the usual relation for capacitors in series. Note again that the same $\sigma$ of free charge appears in computing the net charge on the conducting plates. Thus, the charges on the surface of the dielectrics don't matter in the sense they never enter directly: their effect - which is to lessen the electric field in the materials - is through the introduction of $\epsilon_1$ and $\epsilon_2$.

Sorry I'm a clutz for editing the size of figures...

(Figure credits: Sears and Zemansky's University Physics, by Young and Freedman)

This simplification doesn't work as well with a dielectric because the dielectric is not a conductor which will come to electrostatic equilibrium the same way as the rest of the circuit.

• What do you mean? Are you suggesting the expression $1/C=1/C_1+1/C_2$ is incorrect? – ZeroTheHero Mar 12 '17 at 20:00