Calculating Potential Difference in a Charged Pipe Using Gauss's Law

[gathan77]

This was a recent problem on an exam that I did poorly on that I was wondering if I could acquire some assistance to further my understanding.

Homework Statement

A long plastic pipe has an inner radius A and an outer radius B. Charge is uniformly distributed over the volume A<r<B, and the amount of charge is ρ C/m^3. Find the potential difference between r=B and r=0.

Homework Equations

∫ E dA = q enc / ε

The Attempt at a Solution

I'm stumped on how to carry out this problem, but I'm thinking that using Gauss's Law to find the electric field would be helpful so that you can then integrate it to find V. However, my cylindrical integration skills are a little shaky and I'm not sure if this is the the correct way. Any help would be appreciated. Thanks

 

[Mindscrape]

Yes, Gauss's law is always the way to go when you have symmetry. First off, there is no charge on the inside, so we know that from 0<s<a (I'm going to use s as the variable for circular radius and phi for the angle) will be 0. So all we really need to do is find the electric field inside the plastic. Since no dielectric is given, so I will just use E, but you can use D later if you want. So...

$$\iint_S \mathbf{E} \cdot d\mathbf{a} = \frac{\iiint_V \rho(\mathbf{s}') dV'}{\epsilon_0}$$

Since I think you can probably manage the surface integral, let's focus on the volume element.


Use the proper Jacobian for the cylindrical coordinate

$$\int_0^L \int_0^{2\pi} \int_A^B \rho s' ds' d\phi' dz'$$

Can you take it from here?

$$V = - \int_a^b \mathbf{E} \cdot d \mathbf{l}$$

where the a and b are technically different a and b from the ones in your problem, and might even be because I don't remember which direction the integral goes in.

 

[gathan77]

Ok, I've been working on this a little, so I was wondering if you could tell me if I'm wrong.

E L 2 π r = ρ * L π ( r$$^{2}$$ - a$$^{2}$$ ) / ε

So,

E = ρ ( r$$^{2}$$ - a$$^{2}$$ ) / 2 ε r

Knowing

ΔV = - ∫ E dr
= - ∫ [ρ ( r$$^{2}$$ - a$$^{2}$$) / 2 ε r ] dr
= ( ρ / 2 ε ) ∫ [( a$$^{2}$$ - r$$^{2}$$ ) / r ] dr

Then we can split that into two integrals. So,

ΔV = ( ρ / 2 ε )[∫( a$$^{2}$$ / r ) dr - ∫ r dr]
= ( ρ / 2 ε ) [a$$^{2}$$(ln(a) - ln(b)) - (a$$^{2}$$ / 2 - b$$^{2}$$/2]

Does this look right?

 

[Mindscrape]

Looks good to me. :)