Potential at centre of cylinder

1. Jun 5, 2013

Pranav-Arora

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
I have no idea on this one. I can't understand what happens when some portion of the cylinder is earthed and other is maintained at 110V. The potential of the earthed portion is 0V but how does that even help me?

Any help is appreciated. Thanks!

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2. Jun 5, 2013

voko

Is this not just a simple application of the superposition of electric potentials?

3. Jun 5, 2013

haruspex

Don't you need to know the charge distributions to do that? The OP is a Dirichlet boundary value problem, no?

4. Jun 5, 2013

Pranav-Arora

What is this Dirichlet boundary value problem?

5. Jun 5, 2013

TSny

Suppose you choose an imaginary circle of radius r < R lying in a plane perpendicular to the axis of the cylinder and centered on the axis of the cylinder.

Try to show that the integral ∫0 V(r,θ)dθ is independent of the value of r, where the integral is over the circle of radius r and θ is the polar angle.

If you can do that, then you can relate V on the axis to the values of V on the cylinder.

Last edited: Jun 5, 2013
6. Jun 5, 2013

Pranav-Arora

I understand what you say here but how am I even supposed to find $V(r,\theta)$?

I don't even have the charge distribution, only the potential of two portions of cylinder.

7. Jun 5, 2013

TSny

You don't need to find V(r, θ) in order to show that ∫V(r,θ)dθ is independent of r.

Although not obvious, you can show this by starting with Gauss's law for a closed cylindrical Gaussian surface of radius r as shown. See if you can express the net flux through the surface in terms of an integral of Er, the radial component of the electric field. Then relate Er to V(r,θ) inside the integral.

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8. Jun 5, 2013

Pranav-Arora

Flux through the Gaussian surface is $\phi=E_r\cdot 2\pi rh$ but this does not involve any integral.

I think I cannot consider $E_r$ as constant over the whole Gaussian surface, right?

From Gauss law, $\phi=\int \vec{E_r}\cdot \vec{dA}$. $\displaystyle E_r=-\frac{dV}{dr}$ but this does not involve $\theta$.

9. Jun 5, 2013

TSny

Er cannot be assumed to be constant over the surface. So, can you bring it out of the flux integral?

[EDIT]
Since V is a function of r and θ, ∂V/∂r will also be a function of r and θ.

Last edited: Jun 5, 2013
10. Jun 5, 2013

Pranav-Arora

I don't know how to solve $dV(r,\theta)/dr$.

11. Jun 5, 2013

TSny

You don't need to solve dV/dr. Express the flux integral in terms of ∂V/∂r. The goal is to use the flux integral to show d/dr ∫V(r,θ)dθ = 0, which means ∫V(r,θ)dθ over a circle of radius r is a constant (independent of the size of the circle).

Last edited: Jun 5, 2013
12. Jun 5, 2013

haruspex

See http://physics.usask.ca/~hirose/p812/notes/Ch2.pdf.
If the potential is specified on a closed surface, the potential off the surface is uniquely determined in terms of the surface potential. This is known as Dirichlet's boundary value problem

13. Jun 6, 2013

Pranav-Arora

Do I need to simply substitute $E_r=∂V/∂r$?

That's too much mathematics. Is there no easier way to this problem?

14. Jun 6, 2013

voko

Well, the solution to a Dirichlet problem, involving Green's function, is an application of superposition :)

But you are correct in that it is not simple and not direct as my original suggestion implied. I think I misunderstood the problem initially.

Last edited: Jun 6, 2013
15. Jun 6, 2013

voko

This is the standard approach to such problems. Besides, most of mathematics there is not for you do, but merely to follow. If you follow, you will see in one example that most of what needs to be done for this problem is done there, you just need to plug in your data to get the solution.

16. Jun 6, 2013

TSny

Yes. What does Gauss' law look like for the chosen Gaussian surface after making the substitution $E_r=-∂V/∂r$.

That should bring you to within a few simple steps of the answer to the question.

17. Jun 6, 2013

Pranav-Arora

$\displaystyle \phi=\int -\frac{∂V}{∂r}dA$

But what is $dA$?

18. Jun 6, 2013

TSny

What part of the Gaussian surface are you integrating over here?

19. Jun 6, 2013

Pranav-Arora

Is $dA=rd\theta dy$ where y is the distance of the patch(dA) from the top base of the Gaussian surface?

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20. Jun 6, 2013

haruspex

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