Resistance of a hollow sphere

[TonyMcC]

Ah, but this is an exceptional circuit. One terminal of a battery is connected to the inside of the sphere without needing a hole. The other terminal is connected to the outside of the sphere which is coated with magical material which has no resistance. This will provide the radial current!

On a more serious note I am sure this must just be an exersise in integration.

[swansont]

There is no electric field inside in electrostatics. Current flow means it's a dynamic, rather than static, situation.

 

On a more serious note I am sure this must just be an exersise in integration.

 

Undoubtedly. One could also investigate a frustum-shaped resistor, which would at least be physically realizable.

[rmw]

convert the sphere to an equivalent area elemental sheet, areas of successive sheets increasing with the radius. and make it out of high grade unobtanium, polished to mirror finish so you can see the truth in it.

[TonyMcC]

Can I suggest the question would have been clearer if the units were given for resistivity. For instance resistivity of copper is approx 1.7 micro ohms -cm or 17 nano ohms-m.

[DerekTX]

Okay, this is what I have gotten so far..

 

J=I/A and rho = E/J

So E=rho *(I/4*pi*r^2)

To get the resistance, I'll take the integral of E

 

R = rho * integral of (1/4*pi*r^2)dr

= rho * 1/4pi * ((10^-6)-(10^-2))

= 1.35*10^-11

Your physics is right. Using what you did I got 1.35*10^-3

 

I can see what you're doing wrong, but not whether darkenlighten or I am correct...

 

You did this:

 

[math]R = \rho \times \frac{1}{4\pi} (10^{-6}-10^{-2})[/math]

 

But the expression you wanted to integrate was

 

[math]R = \rho \int \frac{1}{4\pi r^2} \, dr[/math]

 

The r² is in the denominator of the integral.

 

darkenlighten: with the radii in units of meters (10^-2 and 10^-6), I get 10^-3 in my answer. Did you change any other values, or are we doing different integrals?

This is exactly right.

 

 

I just don't understand why. to me, in my mind, the formula ought to be [math]R=rho*L/dA[/math] but then I wind up with what darkenlighten has, which is wrong. I don't understand why the L magically goes away and you integrate the entire [math]1/4pir^2[/math] expression.

 

happy dead thread-aversary.