# Homework Help: Electrodynamics, Need help with problem

1. Oct 9, 2012

### Stan12

1. The problem statement, all variables and given/known data

An infinitely long cylindrical volume of radius R contains a charge density ρ(s)=ks4 where k is a constant and s is the distance from the axis of the cylinder. Note that this is NOT a constant density.

a) Find the electric field everywhere in space.

b) From your result in part a) find the electrostatic potential inside the cylinder, assuming that the potential vanishes along the axis of symmetry, V(s)|[s=0]=0

2. Relevant equations

∫E*dA

closed surface : ∫E*dA = Qenc /εo

Qenc = ∫ρ dτ

3. The attempt at a solution

Qenc = k ∫∫∫ s'^4 s' ds'dz'dθ'

it came out to be 2∏klR^6 / 6 εo

flux = kR^5 / 6εo

I'm not sure this is correct so I cannot continue to part b)

Last edited: Oct 9, 2012
2. Oct 10, 2012

### rude man

You have (apparently) assumed a unit length of your wire. In your calculations I would have assumed an arbitrary length L intead. Why? Because that way you can check your dimensions in all your terms as you proceed in your calculations.

But, for unit length of wire, your integral is correct. And so is your calculation of flux E(s).

So you can go on. Realize of course that E(s) = what you came up with is valid only for 0 <= s <= R.

So potential is defined as zero on the axis. What is the work needed to take a unit test charge from the axis out to the wire surface and then out some more to infinity? (You are moving the test charge in a radial direction).

3. Oct 10, 2012

### schaefera

Why did your flux lose an R in the numerator? Remember that flux= Q/(e_0). It should be your net charge enclosed and not that divided by R.

Edit: Perhaps you wrote flux meaning the field?

In that case, OP did NOT assume unit length- the L's cancel when you set the flux equal to Q/epsilon, because each involves an L. If he had an L left somewhere the units would be off.

4. Oct 10, 2012

### Stan12

So with the E I can find the potential but plugging it to this,
V(r) = -∫E*dr . Thank you guys!!!

5. Oct 10, 2012

### Stan12

I worked it out today and change some notations, I came up with E(r) = kr^5/6εo for 0<= r <= R and E(r) = kR^6/6rεo for R< r.

6. Oct 10, 2012

### rude man

Never mind. I did not see the small "l" in his integral.

Last edited: Oct 10, 2012
7. Oct 10, 2012

### rude man

You are on the right track!