# Resistance of a hollow sphere

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A hollow sphere with an inner radius r=1 um, and an outer radius R=1 cm is made from copper. A current flows radially; for example, from inside toward outside. Find the resistance of the sphere. The resistivity of copper is p=1.7 * 10^-8

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What have you tried so far?

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I honestly have no idea where to start one this one. A couple other students in our class have been trying for hours. We have never seen anything like this, and there are no examples in our book.

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How does resistivity relate to resistance?

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I'm wondering how you get connections to both the inside and outside without making a hole in the hollow sphere. This hole will reduce the amount of copper used and its size and shape will have a small bearing on your answer (assuming the hole is small).

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Don't make up a new problem to solve. Work with the task at hand. The problem statement does not require you to figure out how current reaches the inner surface. Focus on Swansont's question and derive a formula based on the relationship and geometry.

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Further to cypress; could I suggest a thought about units of resistivity and how they come to be what they are.

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Well we are given the equation, R = rho*(L/A). Resistance = Resistivity*(Length/Area). In the examples from our book, the length represents the length of a cylindrical object. This is where I am getting stuck. How would I relate the length to a spherical object?

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Well we are given the equation, R = rho*(L/A). Resistance = Resistivity*(Length/Area). In the examples from our book, the length represents the length of a cylindrical object. This is where I am getting stuck. How would I relate the length to a spherical object?

You can view the sphere as having an area, and the length is the distance the current travels. In this case, the area is not constant, which is why this is a more difficult question.

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I suppose it is ok to give a few more hints. The magic term here is: integration. Cut your object into small pieces that you can handle, let the pieces be really tiny (leetspeak-term: infinitesimal), possibly make some approximations, then put the pieces together to form your structure (which is where adding up contribution from different pieces comes in - note that an integration is just a limit of a summation when the pieces become tiny).

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Indeed. If you consider the current flowing from the inside to the outside, you can consider the inside surface of the conducting sphere to be the area of your conductor. It travels a given length until it reaches the outside. But the area changes as it does so, so integration will be required.

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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

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I tried your solution and it makes sense; however, I don't see how you got $1.35\times10^{-11}$. I got $10^{-3}$. I don't know if I'm right; my calculator is acting up and displaying strange things. Double-check?

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Both of you are probably correct, so you wanna check your units. I converted all length units to meters, which gave me: $1.353\times10^{-7}$.

I think it is easier to go with this logic (since you have pretty much came to the conclusion). First realize what L is: $1cm - 1 \mu m$ and then find the area. Since it's a hollow sphere, its area changes with distance, r, but r is just the length so $R = \rho*(\frac{L}{4*\pi*r^2})$ and then since $r = L$ $\Rightarrow R = \rho*(\frac{1}{4*\pi*L})$

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I'm not sure what I am doing wrong but I keep getting 10^-11.

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type out the numbers you use and the units you are using for length

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I'm not sure what I am doing wrong but I keep getting 10^-11.

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

You did this:

$R = \rho \times \frac{1}{4\pi} (10^{-6}-10^{-2})$

But the expression you wanted to integrate was

$R = \rho \int \frac{1}{4\pi r^2} \, dr$

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?

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I dont think that integration is good logic.

What I would do if you really want to put an integral in there would be to find the area using a double integral $A = \int_{0}^{2\pi} \int_{0}^{\pi} r^2 sin(\theta) d\theta d\phi$

And then realizing that $r = L$ and solving the rest of the equation as stated before. Though doing this integral is really trivial, since the area of a sphere is already derived.

But I used: $R = (\frac{1.7\times 10^{-8}}{4\pi*[(1\times 10^{-2}) - (1 \times 10^{-6})]})$

Edited by darkenlighten
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I dont think that integration is good logic.

Why not?

What I would do if you really want to put an integral in there would be to find the area using a double integral $A = \int_{0}^{2\pi} \int_{0}^{\pi} r^2 sin(\theta) d\theta d\phi$

And then realizing that $r = L$ and solving the rest of the equation as stated before. Though doing this integral is really trivial, since the area of a sphere is already derived.

Right. I'm trying to do the same thing, but with the formula for the surface area of a sphere, rather than using a double integral.

But I used: $R = (\frac{1.7\times 10^{-8}}{4\pi*[(1\times 10^{-2}) - (1 \times 10^{-6})]})$

How does this account for the changing area of the shell?

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Oh, okay. I understand where I was going wrong. Thanks for the correction.

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You guys seem to have cracked it using calculus. I'm sure that must have been the objective of the questioner because of the impossible situation described. For a practical crack at the problem which would give a good approximation you could do the following:- (This is no doubt what your calculus does more accurately). Consider the sphere as having several layers, like an onion. For each layer calculate a resistance, using the average of outer area and inner area(approximation). If you considered 50 layers you would have 50 resistances. Think whether you should consider these resistances as in series or in parallel and make your simple calculation. I suppose this sphere is coated in some magical material which has no resistance in order to make radial current flow?

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That's exactly what calculus does more accurately, because the layers become infinitely thin and there are an infinite number of them.

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I know swansont - you'll have to forgive me. I studied a bit of calculus 50 or so years ago. I am now retired and during my lifetimes work I never had recourse to use it - to say I am rusty would be an understatement! However I did wonder if my basic description of what calculus did in this instance would be of some use to anyone struggling with the concept.

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I know swansont - you'll have to forgive me. I studied a bit of calculus 50 or so years ago. I am now retired and during my lifetimes work I never had recourse to use it - to say I am rusty would be an understatement! However I did wonder if my basic description of what calculus did in this instance would be of some use to anyone struggling with the concept.

Well, since then calculators have gotten better! Now you can use a calculator to solve integration. For example, the integrator. Some actual calculators can do it too.

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The sphere is a hollow charged conductor when connected to one terminal of the circuit. As such, a Faraday cage, there is no electric field inside and therefore no potential gradient within it.

My sensible brain says it must work, but..............

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