Resistance for a sphere in a half sphere containing liquid

In summary, the sphere is immersed in a liquid of resistivity \rho . The resistance between the sphere and the walls of the bath can be estimated using the resistivity equation.
  • #1
TFM
1,026
0

Homework Statement



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

Homework Equations



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

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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  • #2
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]
 
  • #3
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)
 
  • #4
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.
 
  • #5
Could we use spherical polar coordinates, do it for a whole sphere, and then half the value?
 
  • #6
TFM said:

Homework Equations



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


TFM

Surely this should be

R = rho *L/A
 
  • #7
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.
 
  • #8
So I dion't need to use:


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

Expressed in Spherical polar coordinates?
 
  • #9
TFM said:
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.
 
  • #10
Okay, so What should I do for the length part?
 
  • #11
Surely the length is just the difference between the radii?
 
  • #12
Okay, so:

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

Does this look okay?
 
  • #13
TFM said:
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.
 
Last edited:
  • #14
Okay, so it should be:

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

?
 
  • #15
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?
 
  • #16
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.
 
  • #17
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?
 
  • #18
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
 
Last edited:
  • #19
TFM said:
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?

Bingo
why call r tau?
 
  • #20
Okay, so talke the ocnstants outside:

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


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

Makes:

[tex] R = \frac{\rho}{2\pi} [r^{-1}/1]_a^b [/tex]

Thus:

[tex] R = \frac{\rho}{2\pi} [\frac{1}{r}]_a^b [/tex]

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

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

Does this look okay?
 

Related to Resistance for a sphere in a half sphere containing liquid

What is the concept of resistance for a sphere in a half sphere containing liquid?

The concept of resistance for a sphere in a half sphere containing liquid is a measure of the force that opposes the motion of the sphere as it moves through the liquid. It is affected by factors such as the size and shape of the sphere, the density and viscosity of the liquid, and the velocity of the sphere.

How is the resistance of a sphere in a half sphere containing liquid calculated?

The resistance of a sphere in a half sphere containing liquid can be calculated using the Stokes' law formula, which takes into account the radius of the sphere, the viscosity of the liquid, and the velocity of the sphere. Other factors such as the density of the liquid and the shape of the sphere may also be considered in more complex calculations.

What factors affect the resistance of a sphere in a half sphere containing liquid?

The resistance of a sphere in a half sphere containing liquid is affected by several factors, including the size and shape of the sphere, the density and viscosity of the liquid, and the velocity of the sphere. Other factors such as the roughness of the surface of the sphere and the presence of any obstacles in the path of the sphere may also affect the resistance.

What is the significance of resistance for a sphere in a half sphere containing liquid?

The concept of resistance for a sphere in a half sphere containing liquid is important in various fields such as fluid dynamics, hydrodynamics, and aerodynamics. It helps in understanding the behavior of objects moving through fluids and can be useful in designing efficient structures, such as ships and submarines, that need to move through water with minimum resistance.

How can the resistance of a sphere in a half sphere containing liquid be reduced?

The resistance of a sphere in a half sphere containing liquid can be reduced by changing any of the factors that affect it, such as the size and shape of the sphere, the density and viscosity of the liquid, and the velocity of the sphere. For example, using a more streamlined shape for the sphere, using a less dense or less viscous liquid, or reducing the velocity of the sphere can help in reducing the resistance. Other techniques such as adding a lubricant or using special coatings on the surface of the sphere may also be effective in reducing resistance.

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