# Calculate the capacitance

1. Dec 14, 2009

### fluidistic

1. The problem statement, all variables and given/known data
Calculate the capacitance of a spherical capacitor such that its center and up to $$R_2$$ is vacuum.
Then from $$R_2$$ up to $$R_3$$ there's a dielectric material of constant $$\kappa _2$$. Then from $$R_3$$ up to $$R_1$$ there's a material of constant $$\kappa _1$$.

2. Relevant equations
None given.

3. The attempt at a solution

I've tried many things, but then I realized I was lost since it's a 3 dimensional capacitor, which differs from the 2 dimensional problem I was used to.
Q=C/V. Also, $$\varepsilon _1=\kappa _1 \varepsilon _0$$. And it's similar for $$\varepsilon _2$$.
$$V=-\int_a^b \vec E d\vec l=\frac{Q}{C}$$.
I'm having a hard time finding $$-\int_a^b \vec E d\vec l$$.
For the interior material, $$V=E(R_3 -R_2)$$.

I made an attempt to find E : $$kQ\left [ \frac{1}{R_3}-\frac{1}{R_2} \right]$$. Where $$k=\frac{1}{\varepsilon _0 \kappa _2 4\pi}$$.
Hence $$C _2=\frac{4\pi \varepsilon _0 \kappa _2 R_3 R_2}{R_2-R_3}$$.

I realize that $$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$$ because the capacitor is equivalent to 2 capacitors in series. I'd like to know if my result for $$C_2$$, the interior capacitor is right.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 15, 2009

### kuruman

First you need to derive (or look up) the capacitance of a spherical capacitor. The geometry, however, is incomplete. A capacitor, in general, is two conductors in some spatial configuration. Your statement of the problem says nothing about where the conductors are. What are the radii at which you have the conductors in this particular case?

3. Dec 15, 2009

### fluidistic

You're absolutely right, sorry. The conductor plates are at R_1 and R_2.
I've looked in wikipedia, and instead of my $$C _2=\frac{4\pi \varepsilon _0 \kappa _2 R_3 R_2}{R_2-R_3}$$, I should have got a -sign in front of it... so I almost got it, I don't know where I made a mistake though and I'm willing to see where it is. Although I realize I could have made other errors.

4. Dec 15, 2009

### kuruman

So you know what to do now, right? Two capacitors in series, each with a different dielectric.

5. Dec 15, 2009

### fluidistic

Yes sure.

$$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$$.
Assuming my C_2 as my result with a minus sign, I can get C_1 as easily and thus C, which is the total capacitance.
I thank you for your help. (Once again ).