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Electrodynamics question

  • #1

Homework Statement



I'd type this out but there's a bit too much formulae.
http://www.maths.tcd.ie/~tristan/MA3431/Homework_1.pdf

It's problem 2. I'm just wondering if my solution is correct.

Thanks in advance!

Homework Equations





The Attempt at a Solution



In this problem I'm taking Ro = |xo - x'|

The volume in question is charge-free so the charge density, ρ(x'), is zero so the first term on the right hand side of the potential vanishes.

Also, ∂/∂n(1/Ro) = -1/R2o

Substituting this into the potential function gives the required result.


For the second part of 2 we use the divergence theorem (which I wont state here due to my lack of latex skills) as told.

We know that ∂[itex]\Phi[/itex]/∂n = ∇[itex]\Phi[/itex].n and from the definition of the electric field E we end up with -E.n.

When this is used in the divergence theorem we end up with the volume integral of ∇.E which is equal to ρ/ε which vanishes in a charge free volume.

We now have the required expression.


(Haven't gotten to part 3 yet, will be posted soon.)

I know I haven't explained everything in a great way, but it's a lot easier on paper than it is online to write out loads of partials and surface integrals.
 

Answers and Replies

  • #2
vela
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Homework Statement



I'd type this out but there's a bit too much formulae.
http://www.maths.tcd.ie/~tristan/MA3431/Homework_1.pdf

It's problem 2. I'm just wondering if my solution is correct.

Thanks in advance!

Homework Equations





The Attempt at a Solution



In this problem I'm taking Ro = |xo - x'|

The volume in question is charge-free so the charge density, ρ(x'), is zero so the first term on the right hand side of the potential vanishes.

Also, ∂/∂n(1/Ro) = -1/R2o
Perhaps it was just a typo, but the lefthand side should be written
$$\left.\frac{\partial}{\partial n}\left(\frac{1}{R}\right)\right|_{R=R_0}.$$ The way you wrote it, you're differentiating the constant 1/R0, so the lefthand side would be equal to 0.

Other than that, what you've done sounds fine.
 

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