Electric field and Laplace's equation

In summary, the problem at hand is to show that the magnitude of the electric field in a conducting sheet bent into a wedge shape with a certain angle is proportional to r^{(\pi/\theta) - 1}. This can be solved using separation of variables for Laplace's equation in cylindrical coordinates, taking into account the fact that the potential does not have any 'z' dependence. The solution involves using boundary conditions to determine the values of B_{0}, C, and D, which must go to zero at z = 0 for the solution to be valid. The specific boundary conditions for this problem are not explicitly stated, but it can be inferred that they involve the properties of potentials and conductors.
  • #1
Void123
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Homework Statement



I have to show for a conducting sheet bent along one axis into the shape of a wedge, with a certain angle, that the magnitude of the electric field in the bend is proportional to [tex]r^{(\pi/\theta) - 1}[/tex], where theta is the opening angle.


Homework Equations





The Attempt at a Solution



I'm not sure how to treat this problem, is this just separation of variables for Laplace's equation in three dimensions?
 
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  • #2
You can use separation of variables for Laplace's equation in cylindrical coords. You also know the potential won't have any 'z' dependence so that will simplify your work.
 
  • #3
I know what the solution is, but I don't really understand the physical logic behind it.

If [tex]V(s,\phi) = V_{0} + B_{0}Ln s + \Sigma (s^{n}[A cos (n\phi) + B sin (n\phi)] + s^{-n}[C cos (n\phi) + D sin (n\phi)]) [/tex]

Now, I know that at z = 0 this forces [tex]B_{0}, C, and D[/tex] to go to zero (or else there will be infinite terms), but I don't understand necessarily why. What forces us to conclude, considering cylindrical symmetry, that s -> 0?
 
  • #4
Don't forget you have boundary conditions you need to satisfy.
 
  • #5
Okay, that's what I thought. But which boundary conditions are those if they are not specified in the problem?
 
  • #6
They are specified in the problem. What do you know about potentials and conductors.
 

FAQ: Electric field and Laplace's equation

What is an electric field?

An electric field is a physical field that surrounds charged particles and exerts a force on other charged particles within its range. It is a vector quantity, meaning it has both magnitude and direction. The strength of an electric field is determined by the amount of charge present and the distance from the charged particle.

What is Laplace's equation?

Laplace's equation is a second-order partial differential equation that describes the behavior of electric fields in a region with no charge present. It is used to calculate the electric potential at any point in space, given the boundary conditions of the region.

How is Laplace's equation used in science?

Laplace's equation is used in a variety of scientific fields, including physics, engineering, and mathematics. It is an important tool for solving problems involving electric fields, such as determining the potential distribution in a circuit or the behavior of charged particles in a vacuum.

What is the relationship between electric field and potential?

The electric field and potential are closely related. Electric field is the negative gradient of potential, meaning it is the rate of change of potential with respect to distance. In other words, the electric field points in the direction of decreasing potential and its strength is proportional to the rate of change of potential.

How does the presence of charges affect the electric field and Laplace's equation?

The presence of charges affects the electric field and Laplace's equation by creating a non-zero potential and changing the behavior of the electric field. In the presence of charges, the boundary conditions for Laplace's equation are no longer zero, and the resulting electric potential and field will be different than in a region with no charges.

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