Equipotential surfaces

In summary: Expert SummarizerIn summary, the problem involves finding the equipotential surface at the edge of a uniform electric charged disk. The equipotential surfaces are expected to be concentric circles due to the symmetry of the problem. While the displacement of a point charge on these surfaces must be zero, this is not enough to prove that the surfaces are concentric circles. To fully prove this, one must show that the electric field at any point on the surface is perpendicular to the surface, which can be done by taking the gradient of the potential at a point on the surface.
  • #1
wisky40
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0

Homework Statement


Find the equipotential surface at the edge of a uniform electric charged disk ?

Homework Equations




[tex]\nabla^2 V= - \displaystyle \frac{\rho}{\epsilon} [/tex]


[tex]\displaystyle V= \iiint_R \frac{ {\rho}r dr d\phi dz}{4 {\pi} {\epsilon}|\vec r -\vec r'|}[/tex]

The Attempt at a Solution



In the case of Poisson's equation, I would've have to solve for PDEs, so I tried to aviod it.
In the integral, the distance of the disk radius was almost similar to an arbitrary point close to the border line, so that I couldn't use [tex]R>>r[/tex] as aproximation.
However, I decided to use symmetry, so that I thought if the electric field lines go radially inwards or ourwards, depending on the charge, then, the equipotential are concentric circles, but some classmates told me that was not enough. Finally, I tried use the triple integral to prove that a displacement of a point charge on the equipotential surfaces must be zero.

[tex]\int_{\phi_1}^{\phi_2 } q \displaystyle \hat e_\phi \frac{1}{a}\displaystyle \frac{ \partial V}{\partial \phi} \cdot \hat e_\phi a d \phi =0 \Rightarrow[/tex], so basically my proof consisted on:
[tex] \displaystyle \frac{\partial V}{\partial \phi} = 0[/tex], but nobody agree with me.
So, my question is this. Is there something wrong with my reasoning? or is there any other way?
Regards, wisky 40
 
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  • #2


Dear wisky 40,

Thank you for your post. Your reasoning is correct in that the equipotential surfaces for a uniform electric charged disk will be concentric circles. This is due to the symmetry of the problem, as you mentioned.

However, your proof using the triple integral is not complete. While it is true that the displacement of a point charge on the equipotential surfaces must be zero, this is not enough to prove that the equipotential surfaces are concentric circles.

To fully prove this, you would need to show that the electric field at any point on the surface is perpendicular to the surface. This can be done by taking the gradient of the potential at a point on the surface and showing that it is perpendicular to the surface. This would involve using the partial derivative with respect to the radial coordinate, as well as the partial derivative with respect to the angular coordinate, as you have already done.

I hope this helps clarify your reasoning and provides a way to complete your proof. Keep up the good work in your studies of electromagnetism!


 

1. What are equipotential surfaces?

Equipotential surfaces are imaginary surfaces that connect points with equal potential in an electric field. This means that if a charged particle were to move along an equipotential surface, it would experience no change in potential energy.

2. How are equipotential surfaces related to electric fields?

Equipotential surfaces are always perpendicular to the electric field lines. This means that if a particle moves along an equipotential surface, it will also be moving perpendicular to the electric field lines, and therefore will not experience a change in potential energy.

3. How do equipotential surfaces affect the movement of charged particles?

Charged particles will always move from higher potential to lower potential. Therefore, if a particle is moving along an equipotential surface, it will continue to move in the same direction because there is no change in potential energy. However, if a particle moves from an equipotential surface to a different one, it will experience a change in potential energy and will be affected by the electric field.

4. Can equipotential surfaces exist in non-uniform electric fields?

Yes, equipotential surfaces can exist in non-uniform electric fields. However, they will not be evenly spaced and will have varying shapes depending on the strength and direction of the electric field. In this case, the equipotential surfaces will still be perpendicular to the electric field lines.

5. How can equipotential surfaces be visualized?

Equipotential surfaces can be visualized using equipotential mapping or using electric field line diagrams. In equipotential mapping, the potential values at different points are plotted and connected to form equipotential surfaces. In electric field line diagrams, the electric field lines and equipotential surfaces are drawn together to show their relationship. Additionally, there are computer simulations and physical models that can also be used to visualize equipotential surfaces.

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