poisson's equation

[physica]

Sorry for the hibernation recently I have assignments in and this question has had me stuck for 2 days straight. What I get for the first 2 parts of the question is to apply poisson's equation with cylindrical coordinates. Only the radius will change so del squared will only consist of the radius. For the general solution I get the following:

 

V( r )=(p/6EE_0)(r^2)-(A/r)+B

 

I am completely stuck on part C. I think that the boundary conditions are: V(a)=V(2a)=0. I use simultaneous equations to solve for A and B but my result is completely different to the solution given in C. Where do the logs come from? Where have I gone wrong? I would be very grateful if someone could nudge me in the right direction.

 

Many thanks for your time

Edited January 29, 2015 by physica

[physica]

There has been an update, I have realised that I have used spherical coordinates to derive the general solution which is silly. My new general solution is:

 

V( r )=A(Ln[ r ])-[(pr^3)/6EE_0]+C

 

However, my boundary conditions: V(a)=V(2a)=0 make the general solution collapse to zero completely which is definitely wrong.... is anyone out there?

[studiot]

Well poisson's equation in cylindrical coordinates is

 

[math]{\nabla ^2}V = \frac{1}{r}\frac{\partial }{{\partial r}}\left( {r\frac{{\partial V}}{{\partial r}}} \right) + \frac{1}{{{r^2}}}\frac{{{\partial ^2}V}}{{\partial {\theta ^2}}} + \frac{{{\partial ^2}V}}{{\partial {z^2}}} = - \frac{\rho }{\varepsilon }[/math]

 

The reference to a long cylinder provides a value for

 

[math]\frac{{{\partial ^2}V}}{{\partial {z^2}}}[/math]

 

and radial symmetry provides a value for

 

[math]\frac{1}{{{r^2}}}\frac{{{\partial ^2}V}}{{\partial {\theta ^2}}}[/math]

Leaving only the first term and therefore reducing the equation to a solvable ordinary one.

 

Now the physics is that the potential at the boundary with any conductor is a constant.

 

So if we reckon zero potential at the inner surface of the outer conductor as zero and the potential at the outer of the inner one as v (a constant), we can obtain an expression for V®.

 

This leads to the logarithmic ratio solution.

 

 

 

You are then asked to rework with v = 0 in part c.

 

I hope this is enough of a nudge as I'm off out on the town tonight.

Edited January 31, 2015 by studiot

[physica]

Hope you enjoyed your night on the town. Many thanks for the nudge I have now solved it. It turned out that my general solution was wrong. Applying the boundary conditions to my new general solution gave the equation given in the question.