# Homework Help: Coulomb's law and spherical charge distribution

1. Nov 7, 2013

### Avatrin

1. The problem statement, all variables and given/known data
Find the E produced by a spherical charge distribution with uniform charge density at a point inside the sphere, using triple integration.

2. Relevant equations
E = 1/4πε ∫f(x,y,z)/r^2 dV

3. The attempt at a solution
f(x,y,z) = p
Position of point = d = (a,b,c)

E = p/4πε ∫∫∫1/r^2 dxdydz

I tried several things. Among them:

I change to spherical coordinates.
0≤θ≤pi
0≤φ≤2pi
r ≤ R

E = 1/4πε ∫∫∫p/r^2 *r^2sinθdrdθdφ = p/4πε ∫∫∫sinθdrdθdφ = pR/ε

I tried setting (r-d)^2≤R^2, but didn't get anywhere.

Anyhow... The answer I get is not pr/3ε (I dont even know if the r in this answer is supposed to be the radius of the sphere or the distance from origin to what I called d)

2. Nov 7, 2013

### voko

You don't have to position the point at (a, b, c). This just makes things more complicated than it needs to be. Assume that the point is on the x-axis, at distance a from the origin (why can you do this?).

3. Nov 8, 2013

### Avatrin

Because I am dealing with the symmetries of a sphere (and can just turn the sphere around if the point is somewhere else), but I still do not know how to get the right integral.

4. Nov 8, 2013

### voko

Take a small element at $(r, \theta, \phi)$ in the sphere. What is the electric field due to it at $(a, \pi/2, 0)$?

Note the field is vectorial. Due to the symmetries, you can probably figure out the direction of the field without computing it explicitly, so you only have to consider the components of the field in that direction.

5. Nov 8, 2013

### Avatrin

The force is pointing in the radial direction.

E = k * dV/r12
r1 is the distance from dV to a

I tried rewriting r1 in terms of r and a using the law of cosines and got:

r12 = a2 + r2 - 2arcos(π/2 + θ)cos(ϕ)

E*dV = k *dV / (a2 + r2 - 2arcos(π/2 + θ)cos(ϕ)) =

This cannot possibly be the integral they are expecting me to solve (but, then again, I am horrible with spherical coordinates, so maybe there is something I am forgetting).

6. Nov 9, 2013

### voko

I do not think your expression for the square of the distance is correct. In Cartesian coordinates, the square of the distance between $(x, y, z)$ and $(X, Y, Z)$ is $(x - X)^2 + (y - Y)^2 + (z - Z)^2$. In this case, $(X, Y, Z) = (a, 0, 0)$, so the square of the distance is $(x - a)^2 + y^2 + z^2 = a^2 - 2 ax + x^2 + y^2 + z^2 = a^2 - 2 a r \sin \theta \cos \phi + r^2$.

Secondly, $E = k dV /r_1^2$ is the magnitude of the entire vector of the electric field. You need to find its component parallel with the x axis.

Thinking about this some more, I believe it will be more convenient to place the point where the electric field is evaluated on the z-axis, also at distance $a$ from the origin. Then the square of the distance is $a^2 - 2 ar \cos \theta + r^2$, which has no dependency on $\phi$. Then you need to find the component of the electric field in the direction of z.

7. Nov 9, 2013

### Avatrin

You are right. I wrote a2 + r2 - 2arcos(π/2 + θ)cos(ϕ). It should have been a2 + r2 - 2arcos(π/2 - θ)cos(ϕ) (since cos(π/2 - θ) = sin(θ)).

But, if I try spherical coordinates, and try to get only the z-component, I get:

k∫∫∫ cos(θ) r2 sin(θ)/ (a2 + r2[/SUP - 2arcos(θ)) = k∫∫∫ (r2 sin(2θ) / 2(a2 + r2 - 2arcos(θ))) dϕdθdr = 2kpi ∫∫ (r2 sin(2θ) / 2(a2 + r2 - 2arcos(θ))) dθdr

Another integral that seems too complex for this course.

Is there any clever trick I am missing? Or is there no easier way to do this?

8. Nov 9, 2013

### voko

If you think you get the z-component by multiplying the magnitude by $\cos \theta$, you are mistaken.

The E-field is directed along the displacement vector. The displacement vector is $(-x, -y, a - z)$. The unit vector in the direction of the displacement vector is $$\frac 1 {\sqrt{a^2 + r^2 - 2 a r \cos \theta }} (-x, -y, a - z)$$ Hence, the z-component of the field unit vector is $$\frac {a - z} {\sqrt{a^2 + r^2 - 2 a r \cos \theta }} = \frac {a - r \cos \theta } {\sqrt{a^2 + r^2 - 2 a r \cos \theta }}$$ Now that you have the magnitude of the field, and the z-component of the field unit vector, what is the z-component of the field?

9. Nov 11, 2013

### Avatrin

Right.. I always forget that there is supposed to be an r with a hat on it in the equation. In that case I get:

EzdV = kp (a-rcos(θ)) r2 sin(θ)dθdrdϕ/ (a2 + r2 - 2arcos(θ))3/2

Integrating that:

2∏kp∫∫(a-rcos(θ)) r2 sin(θ)dθdr/ (a2 + r2 - 2arcos(θ))3/2

I am again probably doing something wrong. I cant make the integral above become 2r/3.

Also, why is the displacement vector (-x,-y,a-z)?

Last edited: Nov 11, 2013
10. Nov 12, 2013

### voko

The integrals look OK to me now. Try $u = \cos \theta$.

The displacement vector is $(-x, -y, a - z)$ because we evaluate the field due to the element at $(x, y, z)$ at the point $(0, 0, a)$, hence the displacement is $(0, 0, a) - (x, y, z) = (-x, -y, a - z)$.