Difficult Electric field

  • #1

Homework Statement


To find the electric field of a uniformly charged sphere at some distance using the result of the field at an axial of a charged disc.

Homework Equations





The Attempt at a Solution



Well,I approached this proble m in the following way:
You may verify that
E’=(σz’/2ε)[(1/z’)-{1/√(z’^2+r’^2)}is the expression for electric field at an axial point due to the charged disc. We use primed co-ordinates to any such disc located at an arbitrary place in the sphere. z axis is taken vertical, r’ is the radius of a disc.
Now consider the sphere to be made up of such charged discs, in succession; i.e. one placed on another and so on. For a very large number of discs, we may assume contribution from each disc to be dq and in that case, it is easier to think of a volume charge density ρ distributed all over the volume.
Then, dE={(dq/πr’^2)z’/2ε}[1/√(z’^2+r’^2)]
Putting z’=(z-Rcosθ) and r’=Rsinθ in the expression we get a formidable expression. However, dq has not yet been replaced. We Should replace it with ρ dV where dV is the standard volume element in spherical polar co-ordinates.
Now, the problem is how to evaluate the integration. Ф integration is not a problem. Θ integration is bizarre. What about radial co-ordinate integration? I thought only ∫dV will suffice. My friend says that each and every R over here: dq/πr’^2 and over here: [1/√(z’^2+r’^2)] are to be integrated.
I will simply die then.
However, it’s not any homework problem as you could easily see from its level of difficulty. I thought there should exist a way to approach in this path.
 

Answers and Replies

  • #2
siddharth
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We Should replace it with ρ dV where dV is the standard volume element in spherical polar co-ordinates.
I think this is incorrect. Since the infinitesimal charge is contained in the thin disc, the infinitesimal volume should be the volume of the thin disc, and not the standard volume element in spherical polar co-ordinates.
 
  • #3
Then how would you proceed?A thin disc has no volume.And if you think of a pillbpx like thing,then what would be the appropriate volume element?
 
  • #4
siddharth
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Then how would you proceed?A thin disc has no volume.And if you think of a pillbpx like thing,then what would be the appropriate volume element?
The thin disc, or cylinder, should have an infinitesimal height [itex] dh=R d\theta[/itex]. So, the corresponding volume would be [tex]\pi R^2 \sin^2 \theta R d\theta [/tex]
 
  • #5
However,thank you for replying.It's done,not in your way.I used cylindrical co-ordinates all the way,as per the demand of the problem.Use volume element dV=(pi r'^2)dz' and r'+sqrt(R^2-z'^2) and also the distance upto the point from the centre is (z-z').
I think spherical polar co-ordinates will be a little bad choice to do it.The symmetry is not spherical
 
  • #6
siddharth
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However,thank you for replying.It's done,not in your way.I used cylindrical co-ordinates all the way,as per the demand of the problem.Use volume element dV=(pi r'^2)dz' and r'+sqrt(R^2-z'^2) and also the distance upto the point from the centre is (z-z').
I think spherical polar co-ordinates will be a little bad choice to do it.The symmetry is not spherical
Just to clarify, I didn't use the standard volume element in spherical polar coordinates for this question. Since the field of the disc is know, simply sum up the field due to the infintesmally thin disks over the z axis

So, what I did was to reduce the integral to one variable (ie, theta), and integrate over that variable.
 

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