# Electric field due to a Solid Non-Conducting Sphere

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1. Dec 4, 2015

### Siddharth Barua

1. The problem statement, all variables and given/known data
"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.
2. Relevant equations
Coulomb's law for element field dE

3. The attempt at a solution
I integrated the resulting function. I used spherical polar coordinates and have obtained dq, rho and dv

2. Dec 4, 2015

### DuckAmuck

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. Dec 4, 2015

4. Dec 4, 2015

### DuckAmuck

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. Dec 4, 2015

### Siddharth Barua

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. Dec 4, 2015

### Siddharth Barua

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. Dec 5, 2015

### SammyS

Staff Emeritus
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. Dec 5, 2015

Let u=Cosθ

9. Dec 7, 2015

### DuckAmuck

You can use Coulomb's law! You don't even need to do an integral.
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: $$E_{out} = \frac{kQ}{r^2}$$

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: $$Q' = \frac{r^3 Q}{R^3}$$

So now, just plug and chug: $$E_{in} = \frac{kQ'}{r^2}$$
$$E_{in} = \frac{k}{r^2} \frac{r^3 Q}{R^3}$$
$$E_{in} = \frac{kQr}{R^3}$$

10. Dec 7, 2015

### SammyS

Staff Emeritus
I looks like this fact is precisely what OP is to show by doing the full blown integration.

11. Dec 11, 2015

### Siddharth Barua

12. Dec 11, 2015

### haruspex

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. Dec 12, 2015

### J Hann

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. Dec 12, 2015

### SammyS

Staff Emeritus

15. Dec 13, 2015

### J Hann

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