Sphericial Capacitance w/Dielectrics

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The discussion focuses on calculating the capacitance of a spherical capacitor composed of two insulating shells with different dielectric strengths, k1 and k2, and a vacuum gap between metallic shells. The capacitance is derived using the formula C = Q(free) / ΔV, where ΔV is calculated through integrals of the electric field across the different mediums. The integration process involves considering the contributions from each dielectric and the vacuum, leading to the total capacitance equation. Participants confirm that the final expression for capacitance should reflect the reciprocal relationship, indicating that the solution is indeed 1/Ctotal. The calculations and methodology presented are validated by the contributors.
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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.

graph2.jpg


Total C = Q(free) / \DeltaV where \LambdaV is the varied potential between the two metal shells.

\LambdaV = \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?
 
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mitleid said:
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.

graph2.jpg


Total C = Q(free) / \DeltaV where \LambdaV is the varied potential between the two metal shells.

\LambdaV = \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 :

above looks good.

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?

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...
 
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 let's 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...
 
mitleid said:
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 let's 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...

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