Electric field due to a Solid Non-Conducting Sphere

  • #1

Homework Statement


"Find the electric field due to a solid conducting sphere of uniform charge distribution using Coulomb's law and brute force integration only. Use Gauss' Law to verify the result.

Homework Equations


Coulomb's law for element field dE[/B]

The Attempt at a Solution


I integrated the resulting function. I used spherical polar coordinates and have obtained dq, rho and dv
 

Answers and Replies

  • #2
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All the charge will be distributed on the surface of the conductor. So no charge will be inside.
The field outside will look like the source is a point charge. The field inside will be zero.
 
  • #3
Please read the question carefully. It says "solid" sphere
 
  • #4
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So the charge is uniformly distributed throughout the *solid*, instead of the surface? That's not what occurs in conductors, generally. Usually that's with charged insulators.

In either case, then you use Gauss' theorem to show that E increases linearly with radius inside the sphere, and then behaves like a point charge field outside the sphere.

Inside E ~ r
Outside E ~ 1/r^2
 
  • #5
Okay, that's an error on my part. It is definitely a non conducting solid sphere as it has a uniform charge distribution. But my friend @DuckAmuck the parameters of the question clearly states that I need to use the ONLY coulomb's law and Integrate that to deduce the expression. Thanks anyways.
 
  • #6
I need to integrate this

∫dE= k ∫ ρ{(S^2 dS sinθdθdφ(r-Scosθ)}/((S^2 + r^2 - 2rScosθ)^3/2)

how do i go about it?
 
  • #7
SammyS
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I need to integrate this

∫dE= k ∫ ρ{(S^2 dS sinθdθdφ(r-Scosθ)}/((S^2 + r^2 - 2rScosθ)^3/2)

how do i go about it?
You can start by finding the electric field due to a thin disk of radius, a, and uniform charge density, ρ, at a distance of u along the axis of the disk.
 
  • #8
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I need to integrate this

∫dE= k ∫ ρ{(S^2 dS sinθdθdφ(r-Scosθ)}/((S^2 + r^2 - 2rScosθ)^3/2)

how do i go about it?
Let u=Cosθ
 
  • #9
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You can use Coulomb's law! You don't even need to do an integral.:smile:
Let's say the sphere has radius R, and total charge Q.

So if you're outside the sphere (r>R), it's easy, the field is like a point charge: [tex]E_{out} = \frac{kQ}{r^2}[/tex]

If you are inside the sphere (r<R), it's different. The amount of charge you encounter is reduced, since your Gaussian surface doesn't enclose the entire sphere: [tex] Q' = \frac{r^3 Q}{R^3}[/tex]

So now, just plug and chug: [tex]E_{in} = \frac{kQ'}{r^2}[/tex]
[tex] E_{in} = \frac{k}{r^2} \frac{r^3 Q}{R^3}[/tex]
[tex] E_{in} = \frac{kQr}{R^3}[/tex]
 
  • #10
SammyS
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You can use Coulomb's law! You don't even need to do an integral.:smile:
Let's say the sphere has radius R, and total charge Q.

So if you're outside the sphere (r>R), it's easy, the field is like a point charge: [tex]E_{out} = \frac{kQ}{r^2}[/tex]
...
I looks like this fact is precisely what OP is to show by doing the full blown integration.
 
  • #12
haruspex
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You can use Coulomb's law! You don't even need to do an integral.:smile:
Let's say the sphere has radius R, and total charge Q.

So if you're outside the sphere (r>R), it's easy, the field is like a point charge:
Coulomb's Law does not say it is like a point charge. To apply Coulomb's Law you would first have to show it is like a point charge, but the constraints specified in the problem forbid the easy ways of knowing that. Instead, it explicitly requires Siddharth to solve the integral.
Posts #7 and #8 look appropriate, and probably equivalent.
 
  • #13
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If you can show that the field due to a spherical shell at a point outside the shell is the same as if
all of the charge where concentrated at the center of the shell, then since a solid sphere is composed
of an infinite number of such shells; then for a uniformly charged sphere the field outside the
sphere must be the same as that as for a single charge at the center of the sphere.
 
  • #14
SammyS
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If you can show that the field due to a spherical shell at a point outside the shell is the same as if
all of the charge where concentrated at the center of the shell, then since a solid sphere is composed
of an infinite number of such shells; then for a uniformly charged sphere the field outside the
sphere must be the same as that as for a single charge at the center of the sphere.
What's your point?
 
  • #15
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I posted this because I don't know the difficulty of integrating the field due to charged disks.
Finding the field due to a charged shell is tricky but not difficult.
R = radius of charged sphere
r = radius of sphere centered at point P, the point outside of the charged sphere.
Consider the intersection of the two spheres.
L = distance between centers
A = angle to intersection of spheres (along line of centers) for sphere R
B = angle to intersection of spheres for sphere r
Width of strip on R = R dA
Area of strip = 2 pi R^2 sin A dA
dQ = o 2 pi R^2 sin A dA where o is the charge density of the strip
dE = K dQ cos B / L^2 since components of field perpendicular to L cancel
Now you need to use the Law of Cosines to eliminate some variables and get an integral in r only
r^2 = R^2 + L^2 - 2 R L cos A
2 r dr = 2 R L sin A getting rid of the sine
Also, cos B = (r^2 + L^2 - R^2) / (2 R L) getting rid of the cosine
Now the equation is integrable in r from L - R to L + R
And the charge on the spherical shell is Q = 4 pi o R^2
You should get E = K Q / L^2
 

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