Classic Hemispherical Analyzer

[rbrayana123]

Homework Statement

http://oi50.tinypic.com/4i35g.jpg

I remember learning about this in Chemistry but boy I did not expect to actually prove it works haha. I've actually solved the problem but I feel a little uncomfortable with certain aspects.

Homework Equations

E = \DeltaVq

For uniform circular motion:
a = v^2/r

For two concentric cylinders with equal but opposite charge, the potential difference can be calculated out via Gaussian surface to be:

2kQ/L * ln(b/a) where L is the length of the cylinder.

The Attempt at a Solution

First, I checked the potential difference and found it to be 2V * ln(b/a) which immediately tells me a lot about the system. Therefore, kQ/L = V (units seem to work out).

Conservation of Energy:

Vq = 0.5mv^2 --> v^2 = 2Vq/m

Using Gaussian surfaces, E(r) = 2kQ/Lr = 2V/r.
F(r) = E(r)q = 2Vq/r
a(r) = 2Vq/mr = v^2/r which is the equation of motion for uniform circular motion of radius r.

What's bothering me is that the reason I was able to solve this problem is because I'm *very* familiar with the coaxial cable conductor and quickly recognized it.

Provided the inner and the outer conductor did not have equal but opposite charges, would the potential still be the same or would it somehow be different? I think I solved this problem operating under the assumption they do but is it necessary to make that assumption?

 

[rude man]

Homework Statement

http://oi50.tinypic.com/4i35g.jpg

I remember learning about this in Chemistry but boy I did not expect to actually prove it works haha. I've actually solved the problem but I feel a little uncomfortable with certain aspects.

Homework Equations

E = \DeltaVq

This E is energy but later you used E for electric field ... ? Anyway, energy is not relevant here.

For uniform circular motion:
a = v^2/r

For two concentric cylinders with equal but opposite charge, the potential difference can be calculated out via Gaussian surface to be:

2kQ/L * ln(b/a) where L is the length of the cylinder.

That is correct.

The Attempt at a Solution

First, I checked the potential difference and found it to be 2V * ln(b/a) which immediately tells me a lot about the system. Therefore, kQ/L = V (units seem to work out).

What the ...? What is your V? It's not the potential difference between the two shells and it's not V₀ either ... I am confused ...

Conservation of Energy:

Vq = 0.5mv^2 --> v^2 = 2Vq/m

This V must be V₀.

Using Gaussian surfaces, E(r) = 2kQ/Lr = 2V/r.

Again, I don't know what your V is.

At this point I give up ... you need to get your symbols straightened out.

I think you're on the right track, though. Get the E field as E(Q,r), change that to E(V,r) where V is the pot, diff. betw. the two shells. V is of course -∫_a^bE(r)dr.

Then get E(V,r₀), equate qE(V,r₀) to mv²/r₀, realize (as you kind of did) that qV₀ = mv²/2, and solve for V.

Note that the given answer, 2V₀*ln(b/r₀) - 2V₀*ln(a/r₀) can be more simply written as 2V₀*ln(b/a).