Given: Consider a capacitor connected to a battery of voltage V . Let the capacitor have an area A, and a distance L between the plates. Assume that the capacitor has a layer of linear dielectric (of dielectric constant κ, so that ε = κε0) of thickness L/2 on the lower plate.

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I found the capacitance to be $C=\frac{2\kappa\epsilon_0 A}{L}$. Was I correct about the separation distance? In more common examples of a capacitor, they claim that the plates are flat with nothing in between. In this case, the linear dielectric is as tall as half the total distance between the plates. Am I correct in using the remaining free space, namely $L/2$, as the separation distance in the equation? For reference, the equation is:

$$C=\frac{\kappa \epsilon_0 A}{d}$$


Instant check that shows your answer is wrong: set $\kappa = 1$ as if there is no dielectric, and you don't recover the parallel plate formula for a gap of $L$.

How to solve this problem:

Let the voltage across the capacitor be $V$ and the voltages across the top and bottom be $V_t$ and $V_b$ respectively, then $$V=V_t+V_b,$$ which implies that $$1/C=1/C_t+1/C_b$$ since $V=Q/C$.

Thus essentially this system consists of two capacitors in series. One capacitor is the top gap, and one is the bottom part. Use the formula for plate capacitors twice, with the separation being $L/2$ in both cases, then get the total capacitance by using the formula for two capacitors in series. Some algebra is involved.

You can check the answer you get, again, by setting $\kappa=1$, and seeing if you recover the usual plate formula.

  • $\begingroup$ Why do you say there are 2 capacitors in series? The problem states there is only 1. $\endgroup$ – whatwhatwhat Oct 20 '16 at 23:35
  • $\begingroup$ @whatwhatwhat - think of the bottom part and the top part as two capacitors as a way to solve the problem. $\endgroup$ – Suzu Hirose Oct 20 '16 at 23:36
  • $\begingroup$ So then for the top section, there is a buildup of positive charge on the top plate and negative charge on the surface of the dielectric? $\endgroup$ – whatwhatwhat Oct 20 '16 at 23:42
  • $\begingroup$ @whatwhatwhat - sorry my explanation was not very clear. There is an electric field between the two plates of the capacitor generated by the charges on the plates. The capacitance of this system of two plates is got by considering that the sum of the voltage over the two plates must be equal to the total voltage across the whole capacitor. I will edit the answer to make it clearer. $\endgroup$ – Suzu Hirose Oct 20 '16 at 23:48
  • $\begingroup$ Suzu Hirose is completely right! $\endgroup$ – freecharly Oct 21 '16 at 0:12

The $\mathbf{D}$ field between the plates is (as usual, neglecting fringing effects) uniform and normal to the plane of the plates (assume in the z direction for simplicity) and is given by

$$\mathbf{D} = \frac{Q}{A}\hat{\mathbf{z}}$$

where the plates have charge $Q$ and $-Q$ respectively. The electric field (between the plates) within the dielectric is

$$\mathbf{E_\kappa} = \frac{\mathbf{D}}{\kappa \epsilon_0}$$

and the electric field (between the plates) in air is

$$\mathbf{E} = \frac{\mathbf{D}}{\epsilon_0}$$

Thus, the potential difference between the plates is

$$V = \left(\frac{L}{2}\mathbf{E_\kappa} + \frac{L}{2}\mathbf{E}\right)\cdot\hat{\mathbf{z}} = \left(\frac{L}{2}\frac{\mathbf{D}}{\kappa \epsilon_0} + \frac{L}{2}\frac{\mathbf{D}}{\epsilon_0}\right)\cdot\hat{\mathbf{z}} = \left(\frac{L}{2}\frac{Q}{A\kappa\epsilon_0} + \frac{L}{2}\frac{Q}{A\epsilon_0} \right)$$

The capacitance is then

$$C = \frac{Q}{V} = \frac{1}{\frac{L/2}{A\kappa\epsilon_0} + \frac{L/2}{A\epsilon_0}} = \frac{1}{\frac{1}{C_1}+ \frac{1}{C_2}}$$

which is the formula for series connected capacitors.

  • $\begingroup$ Better explanation than my answer of why it is like two capacitors in series. $\endgroup$ – Suzu Hirose Oct 21 '16 at 1:22
  • $\begingroup$ So I'm still treating this as 1 capacitor correct? My friend was helping with this and she said that one way to solve the problem is to treat the system as 2 capacitors, but I couldn't visualize where the 1st capacitor begins/ends and likewise for the 2nd one. $\endgroup$ – whatwhatwhat Oct 22 '16 at 18:59
  • $\begingroup$ @whatwhatwhat, correct, it is just one capacitor however, it is instructive and intuitively 'nice' that we would (ideally) get the same result if we found the equivalent capacitance of two series connected capacitors with the same plate area $A$ and spacing $L/2$ but with different dielectrics; one with $\epsilon_0$ and the other with $\kappa \epsilon_0$ $\endgroup$ – Alfred Centauri Oct 22 '16 at 22:46

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