Sphericial Capacitance w/Dielectrics

1. Oct 31, 2007

mitleid

A spherical capacitor is made of two insulating spherical shells with different dielectric strengths, k1 and k2, situated between two spherical metallic shells and separated by a vacuum gap. Calculate the capacitance of this system.

Total C = Q(free) / $$\Delta$$V where $$\Lambda$$V is the varied potential between the two metal shells.

$$\Lambda$$V = $$\int^{a}_{d}$$E(r)dr = - $$\int^{a}_{b}$$[(1/4*pi*Eps) * (1/k1)*(Qfree / r^2) * dr] - $$\int^{b}_{c}$$[(1/4*pi*Eps) * (Qfree / r^2) * dr] - $$\int^{c}_{d}$$[(1/4*pi*Eps) * (1/k2) * (Qfree / r^2) * dr]

Q free will be negated in the final equation, since C = Q/V and V includes Q in its numerator (hope that makes sense). So after integrating V through each of the mediums, the two dielectrics and the vacuum, I get the sum :

Ctotal = = Ke[C(dielectric 1) + C(vacuum) + C(dielectric 2) = Ke[(1/k1)*(1/a - 1/b) + (1/b - 1/c) + (1/k2)*(1/c - 1/d)]

Is this correct?

2. Oct 31, 2007

learningphysics

above looks good.

how did you get this part? stick to what you were doing above... get delta V... then take Q/delta V... don't take any shortcuts...

3. Oct 31, 2007

mitleid

I did this next...

delV = - (Qfree/(4*pi*Eps)) [ $$\int^{a}_{b}$$ (dr/(k1*r^2)) + $$\int^{b}_{c}$$ (dr/r^2) + $$\int^{c}_{d}$$ (dr/(k2*r^2))

The general form of the solution for int(dr/r^2) is -(1/r). This lets the pre-existing negative cancel out, and I just have to substitute to get the expression I have above... I've got it written down, just didn't want to type it all out. :)

Also, the solution I came up with should be equal to 1/Ctot, not just Ctot. I think that's right...

4. Oct 31, 2007

learningphysics

Yes, exactly that was 1/Ctot. everything looks good.