Electric Potential

Question

Part a is as follows:

A long cylindrical insulator has a uniform charge density of 1.89 μC/m3 and a radius of 5 cm. What is the electric field inside the insulator at a distance of 4 cm? Answer in units of N/C.

 

I answered part a correctly using Gauss's Law. But now, I'm stuck on part b:

How much work must you do to bring a q = 0.032 μC test charge from 11 cm to 4 cm? Answer in units of J.

 

I figured I would need to integrate from 11 to 4 cm for the electric field and then multiply by q, but I'm a little confused. Can I integrate over kQ/r^2 from 11 to 4cm, then multiply by q?

Answer 1

W = q∫_0.11^0.04 E·dr

E is radial, so E·dr = Edr

Are you working this as a multiregion problem?

Gauss' law:

In the region from r = 0 to r = 0.05 m (inside the cylinder),

E(r) (2πr) = ρ/ε₀ (πr²)

E(r) = ρr / 2ε₀

In the region r > 0.05 m (outside the cylinder),

E(r) (2πr) = ρ/ε₀ π(0.05 m)²

 

E(r) = ρ/2ε₀r (0.05 m)²

W = q∫_0.11^0.05 [ρ/2ε₀r (0.05 m)²] dr + q∫_0.05^0.04 [ρr / 2ε₀] dr

Note you have to break up the work integral into 2 regions, inside and outside the cylinder, because the field has different forms inside and outside the cylinder. Let me know if you need further explanation.

Answer 2

Hi Amy!

 

To supplement Arturo's answer, based on what you wrote above, I think your idea of evaluating the path integral of the electric field is right on.  But you have to use the correct electric field expression you determined in part a, and the correct electric field outside the cylinder, to get the correct result.  kQ/r² is the electric field of a point charge, and so not the correct expression to use in this context.  Arturo's solution shows the way.  Just let him (or any of us) know if you have any questions.