Homework Help: 2 Dielectrics in a Parallel Plate Capacitor

1. Aug 2, 2010

Bobbert

So, I know what happens with the first two cases, but what if the dielectrics are on a diagonal?

2. Aug 2, 2010

Gerenuk

Nice drawing :)

You could try to split this capacitor into pieces. So you make horizontal cuts and for each strip you cut at the boundaries of dielectrics.
Now you use the equation
$$C=\frac{\varepsilon A}{d}$$
for each little piece and add up these partial capacitance according to your equation to yield the overall capacitance.
What you get in the end is
$$C=\frac{A}{d}\frac{\ln\frac{\varepsilon_2}{\varepsilon_1}}{\frac{1}{\varepsilon_1}-\frac{1}{\varepsilon_2}}$$
The only problem is that now its ill-defined what C1 and C2 were. I mean in you first two examples you just remove one colour and stick the plates to whatever is left to get C1 and C2. But if you do that for your diagonal case, then the plates touch and thus are not a capacitor anymore.

Is that clear?

3. Aug 2, 2010

Bobbert

Thanks. The logic / ideas make sense but I am having trouble getting your formula. Could you show an extra step or two?

4. Aug 2, 2010

Gerenuk

One strip has capacitance
$$\frac{1}{dC}=\frac{1}{\frac{dx\cdot dy \varepsilon_1}{h}}+\frac{1}{\frac{dx\cdot dy \varepsilon_2}{d-h}}$$
where x is the distance of one strip from the top, y the distance into the plane, so dx dy is an area element, epsilons are the dielectric constants, d the total separation between plates and $h=\alpha x$ (giving a diagonal) the width of one strip from a single dielectric. Also alpha is adjusted to give $d=\alpha x_\text{end}$. You plug in h and integrate
$$C=\int_0^{x_\text{end}}dx\int_0^{y_\text{end}}dy dC$$.

5. Aug 2, 2010

Gerenuk

Btw, it should be
1/C=1/C1+1/C2 ;)