Power draw of a spherical conductor surrounded by an insulator & a conducting shell

1. Oct 5, 2011

WJSwanson

1. The problem statement, all variables and given/known data

A spherical conductor of radius a is surrounded by a spherical conducting shell of radius b, and the gap is filled with an insulating material of resistivity ρ. A thin wire connects the inner surface of the shell to the surface of the conductive sphere, and a potential of V is applied to the outer surface of the conducting shell.

I. Determine the current drawn from the voltage source.
II. Integrate the power density (σE2) over the insulator volume v and compare to the power drawn from the voltage source.

2. Relevant equations

dR = ρ dr / 4πr2
i = V / R
E = ρ J = -$\partial$V/$\partial$r
Pdissipative = I2R
P/v = σE2
V = -∫E dot dr

3. The attempt at a solution

Part I was fairly straightforward. I found R by integrating from b to a (because the current travels inward from b) and got

R = ρ (b - a) / 4πab

which yielded

i = 4πVab / ρ(b - a)

Part II is giving me fits, however. The power drawn from the voltage source is just I*V. To get the power from the power density σE2 I would need to find an expression for E. Since E is in the same direction as the radial displacement vector r,

V = -∫E dr = -Er, evaluated over the limits of integration (from b to a, in this case).

I have a feeling that this is the route I need to take, but I'm not sure where to take it from there. I also considered using

E = ρJ = ρ(i/A)

but am also not sure where to take that... I figure I would need to find a function J(r) since the current density is r-dependent. (Current is constant over an r-dependent geometry.)
Anyone who can help me with this, or give a hint as to the right direction, would be doing me a huge favor.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 6, 2011

WJSwanson

Re: Power draw of a spherical conductor surrounded by an insulator & a conducting she

Bump for the morning crowd.

3. Oct 7, 2011

WJSwanson

Re: Power draw of a spherical conductor surrounded by an insulator & a conducting she

Okay, so I think I've made some progress, and I'd really appreciate if someone could check my thought process here. Some notes I've made and thoughts I've had since I posted the thread:
• I'll also use U to represent volume, since V is already taken and there's no potential-energy function at play here.
• $J = di/dU$
• $E = \rho J$

Since I found earlier that $dR = dr/4\pi r^{2}$, we have

$i = V/R = 4\pi V ab/\rho(b-a)$

and

$di/dr = 4\pi\sigma Vr$

So now we can use $J = di/dU = \frac{4\pi\sigma Vr dr}{4\pi r^{2} dr} = \sigma V/r$

and thus,

$E = J/\sigma = \sigma V/\sigma r = V/r$ $\Rightarrow$ $E^{2} = V^{2}/r^{2}$

which we sub into the equation

$dP/dU = \sigma E^{2} \Rightarrow dP = \sigma E^{2} dU = V^{2}/r^{2} * \sigma * 4 \pi r^{2} dr = 4 \pi\sigma V^{2} dr$

Integrating dP with respect to r, we conclude that

$P = \int ^{b}_{a} 4\pi\sigma V^{2} dr = 4\pi\sigma V^{2}r |^{b}_{a}$

When evaluated, this yields

$P = 4\pi\sigma V^{2}(b - a)$

as our answer for the integral of the power-density function over the insulator volume.

To compare to the power that is dissipated as heat by the insulator, we simply need to take a straightforward approach:

$dP = V di \Rightarrow P = \int^{b}_{a} V di = \int^{b}_{a} 4\pi\sigma V^{2} dr = 4\pi\sigma Vr |^{b}_{a}$

When evaluated, this yields

$P = 4\pi\sigma V^{2}(b - a)$

demonstrating that the power expenditure is equivalent using either method of calculation.

So what do you folks think? Did I do it right? Any glaring errors I should know about? :)

Last edited: Oct 7, 2011
4. Oct 8, 2011

Redbelly98

Staff Emeritus
Re: Power draw of a spherical conductor surrounded by an insulator & a conducting she

I agree with your result for i, but disagree with your power result. For starters, the "straightforward result" would simply use P=iV, using the expression you already got for i. (No integration required.)

I think you went awry in coming up with an expression like

$$\frac{di}{dr} = 4 \pi \sigma Vr$$

as if i were a function that changes with r. But i is actually a constant expression.

I like that approach better. A would be the surface area of a sphere of radius r that the current is flowing through. That would make E2=____?, then you can integrate the power density over the entire volume between r=a and r=b.

5. Oct 8, 2011

WJSwanson

Re: Power draw of a spherical conductor surrounded by an insulator & a conducting she

Oh. Duh. Thanks, I have this habit of assuming everything is supposed to be a screwball and whatnot because my professor has a proclivity for that kind of thing. And so I overcomplicate things sometimes. :(

6. Oct 8, 2011

WJSwanson

Re: Power draw of a spherical conductor surrounded by an insulator & a conducting she

So for clarity's sake, whatever i is, I treat it as a constant instead of trying to find a differential charge, and set A = the area at an arbitrary radial distance r and just end up dividing i/4pi r^2 to get J?

7. Oct 8, 2011

Redbelly98

Staff Emeritus
Re: Power draw of a spherical conductor surrounded by an insulator & a conducting she

Yes, correct.

8. Oct 8, 2011

WJSwanson

Re: Power draw of a spherical conductor surrounded by an insulator & a conducting she

Thanks so much for that. So from there I would just take E (= $\rho$J), square it, multiply by sigma, and evaluate the integral

P = ∫$\sigma$E2*4$\pi$r2dr

leaving the a's and b's in the current function as a's and b's instead of being silly and trying to find them as functions of r.

Right?