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Resistance for a sphere in a half sphere containing liquid

by TFM
Tags: liquid, resistance, sphere
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TFM
#1
Feb16-09, 02:50 PM
P: 1,031
1. The problem statement, all variables and given/known data

A small copper sphere of radius a is half immersed at the surface of a poorly conducting liquid of resistivity [tex] \rho [/tex]. The liquid is contained in a spherical copper bath of radius b, where b>>a and where the bath is concentric with the sphere. Estimate the electrical resistance between the sphere and the walls of the bath, assuming that the copper can be assumed to be a perfect electrical conductor

2. Relevant equations

[tex] R = \rho \frac{A}{l} [/tex]

3. The attempt at a solution

See I think I need to use the resistivity equation I have above, however, this is only for a wire. Is there a version for situations such as these?

TFM
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merryjman
#2
Feb16-09, 03:25 PM
P: 183
It seems to me that you've sort of got a wire here, it's just that its cross-sectional area changes and it's got a weird shape. If b>>a, then doesn't that really mean that the liquid is shaped like a hemisphere? If so, then I think you can just integrate [tex]\rho dA / l [/tex]
TFM
#3
Feb16-09, 03:37 PM
P: 1,031
That makes sense. So:

[tex] R v= \rho \frac{dA}{l} [/tex]

But what would the limits be? Would it be from 0 to Circle(r=a) - Circle (r=a)

merryjman
#4
Feb16-09, 09:23 PM
P: 183
Resistance for a sphere in a half sphere containing liquid

I agree with zero as the lower limit, this would be the bottom of the pool. The disk's radius at the top would just be b; I think the problem is implying the sphere is small enough to be treated like a point. Dig out the Calculus textbook and you ought to be able to find an example like this one where you're integrating by adding a bunch of disks together.
TFM
#5
Feb17-09, 05:57 AM
P: 1,031
Could we use spherical polar coordinates, do it for a whole sphere, and then half the value?
davieddy
#6
Feb17-09, 07:00 AM
P: 181
Quote Quote by TFM View Post
2. Relevant equations

[tex] R = \rho \frac{A}{l} [/tex]


TFM
Surely this should be

R = rho *L/A
davieddy
#7
Feb17-09, 07:20 AM
P: 181
R is the integral from a to b of (rho/ 2*pi*r^2)dr

Don't understand the relevance of b>>a though, except that you could
make b infinity.
TFM
#8
Feb17-09, 04:15 PM
P: 1,031
So I dion't need to use:


[tex] R = \rho \frac{l}{A} [/tex]

Expressed in Spherical polar coordinates?
davieddy
#9
Feb17-09, 08:43 PM
P: 181
Quote Quote by TFM View Post
So I dion't need to use:


[tex] R = \rho \frac{l}{A} [/tex]

Expressed in Spherical polar coordinates?
r(radius) is a "polar" coordinate. 2*pi*r^2 is the area of a hemisphere.
TFM
#10
Feb18-09, 10:33 AM
P: 1,031
Okay, so What should I do for the length part?
Vuldoraq
#11
Feb18-09, 01:12 PM
P: 272
Surely the length is just the difference between the radii?
TFM
#12
Feb18-09, 01:42 PM
P: 1,031
Okay, so:

[tex] R = \rho \frac{b-a}{2\pi r^2}dr [/tex]

Does this look okay?
Vuldoraq
#13
Feb18-09, 02:33 PM
P: 272
Quote Quote by TFM View Post
Okay, so:

[tex] R = \rho \frac{b-a}{2\pi r^2}dr [/tex]

Does this look okay?
Edit: Sorry (had a beer) ignore what I just said (now deleted). The equation looks good in terms of units, beyond that I am unsure.

Edit 2: Now I feel really stupid. The equation won't work because for a start you would get a negative resistance. I am going to go away now to be alone with my shame. Sorry for not being much use.
TFM
#14
Feb18-09, 02:36 PM
P: 1,031
Okay, so it should be:

[tex] R = \rho \frac{b-a}{2\pi r dr} [/tex]

?
Vuldoraq
#15
Feb19-09, 09:20 AM
P: 272
Okay here is what I think is going on. You have a formula for the resistance, but this depends upon the path length and the cross-sectional area. In this problem neither of those things are constant. You have to consider an infinitesimal path length and area. If we switch to sperical polars you can see that the only factor changing the surface area is r (theta and phi are the same for whatever hemisphere you choose). I think the expression should look something like this,

[tex]\int{RdA}=\int{\rho*dl}[/tex]

Although if I am wrong please could someone correct me?
badphysicist
#16
Feb19-09, 02:44 PM
P: 80
L(or l) should be along the path that the electricity will flow. We need to break this up into infinitesimally thin sheets of thickness dr (the L) with changing area. Thus, you get [tex]R=\int_{a}^{b} \frac{\rho dr}{A}[/tex] where A (the area a hemisphere of radius r) is a function of r and thus you can integrate.
TFM
#17
Feb19-09, 02:52 PM
P: 1,031
Okay, so:

[tex] R=\int_{a}^{b} \frac{\rho dr}{A} [/tex]

We also know the area of a hemisphere is:

[tex] 2\pi r^2 [/tex]

Thus giving:

[tex] R=\int_{a}^{b} \frac{\rho dr}{2\pi r^2} [/tex]

So now would I have to integrate this?
davieddy
#18
Feb20-09, 12:13 AM
P: 181
We are making a pig's ear of solving this.

The current I flows radially.

dV/dr = - resistivity * current per unit area = -rho * I/(2*pi*r^2)

Integrate from a to b to find V.

R = V/I


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