# Laplace equation on two semi-infinite conductor planes

• carloscariell
In summary: Is it possible that the potential would be infinite as r approaches infinity? I don't think so, but I am not absolutely sure. I think that you should be able to argue that the potential cannot be infinite as r goes to infinity (when the potential is bounded at the origin).In summary, two semi-infinite conductor planes with an angle β at a constant potential are being considered. The Laplace equation in polar coordinates is being used to solve for the potential. After some attempts at solving the equations, it is found that the possible solutions are ψ(θ) = A cos λθ + B sin λθ and ϕ(r) = r^λ. However, the solution ϕ(r) = r^-λ must
carloscariell

## Homework Statement

Two semi-infinite conductor planes (like this ∠ ) have an angle β at a constant potential.

Whats the potencial close to the origin?

## Homework Equations

∇²V = (1/r)(∂/∂r){ r ( ∂V/∂r ) } + (1/r²)(∂²V/∂$\theta$²)

## The Attempt at a Solution

Trying for laplacian equation on polar coordinates. But I can't solve-it?

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Show us what you tried.

You might try a separation-of-variables type solution: $V = R(r)\Phi(\alpha)$ and determine separate differential equations for $R(r)$ and $\Phi(\alpha)$.

After some brain spanking, got to this:

Starting with u(r, θ) = ϕ(r) ψ(θ) and putting it in the Laplace Equation(polar)

ϕ''(r) ψ(θ) + (1/r) ϕ'(r) ψ(θ) + (1/r²)ϕ(r) ψ''(θ) = 0

Dividing by (1/r²) ϕ(r)ψ(θ) , I have:

r² ϕ''(r) ϕ(r) + rϕ'(r)ϕ(r) = ψ''(θ)ψ(θ) = λ

Then:

* r² ϕ''(r) + rϕ'(r) − λϕ(r) = 0 (*)
* ψ''(θ) + λψ(θ) = 0 (**)

I think the (**) comes to ψ(θ) = A cos µθ + B sin µθ.

Is it correct until here?

The (*) equation, I failed to solve yet...

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carloscariell said:
* r² ϕ''(r) + rϕ'(r) − λϕ(r) = 0 (*)
* ψ''(θ) + λψ(θ) = 0 (**)

I think the (**) comes to ψ(θ) = A cos µθ + B sin µθ.

Is it correct until here?
Look's good. How is $\mu$ related to λ?
The (*) equation, I failed to solve yet...

Try a solution of the form $\phi(r) = r^{\alpha}$ and see if it will solve the equation for appropriate values of $\alpha$.

TSny said:
Look's good. How is $\mu$ related to λ?

Try a solution of the form $\phi(r) = r^{\alpha}$ and see if it will solve the equation for appropriate values of $\alpha$.

I've forgot to write this: λ = µ²

I''ll try to solve using your suggestion. I'll post it here as soon as I can...

Thanks!

Lol.. it was very easy to solve the ODE... I think I can re write the last equations to:

* r² ϕ''(r) + rϕ'(r) − λ²ϕ(r) = 0 (*)
* ψ''(θ) + λ²ψ(θ) = 0 (**)

So, I'll have

ψ(θ) = A cos λθ + B sin λθ
ϕ(r) = r$^{λ}$

My solution, so, would be V(r,θ) = r$^{λ}$ ( A cos λθ + B sin λθ )

The boundaries are

V(r, 0) = V(r,$\alpha$) = V$_{0}$

But I think I need one more... Can u tell me which is?

carloscariell said:
ψ(θ) = A cos λθ + B sin λθ
ϕ(r) = r$^{λ}$

My solution, so, would be V(r,θ) = r$^{λ}$ ( A cos λθ + B sin λθ )

The boundaries are

V(r, 0) = V(r,$\alpha$) = V$_{0}$

But I think I need one more... Can u tell me which is?

You should have found that ϕ(r) = r$^{-λ}$ would also be a solution. You should be able to argue that this solution has to be thrown out. (Maybe you've already done that.)

For there to be a unique solution for V, I think there would have to be an additional boundary condition that specifies the behavior of V for large r. With the two boundary conditions that you have stated, you should nevertheless be able to find an approximate expression for V near the origin. The expression will have an undetermined constant, but it will at least show you the behavior of V near the origin. The undetermined constant is due to not knowing the boundary condition for V for large r.

TSny said:
You should have found that ϕ(r) = r$^{-λ}$ would also be a solution. You should be able to argue that this solution has to be thrown out. (Maybe you've already done that.)

For there to be a unique solution for V, I think there would have to be an additional boundary condition that specifies the behavior of V for large r. With the two boundary conditions that you have stated, you should nevertheless be able to find an approximate expression for V near the origin. The expression will have an undetermined constant, but it will at least show you the behavior of V near the origin. The undetermined constant is due to not knowing the boundary condition for V for large r.

Yes. I have just find out that I should have ϕ(r) = r$^{−λ}$.

Thanks again!PS: I need to finish this and other questions in 1 hour at least. Back to work!

Well, I overlooked another possible solution to the differential equation for ϕ(r) when λ happens to be zero. This solution is not of the form $r^{\alpha}$, but when you find it you will be able to give a reason for discounting this solution.

You're right. I'll will impose that lim$_{r→∞}$ V(r,$\theta$) = 0

carloscariell said:
You're right. I'll will impose that lim$_{r→∞}$ V(r,$\theta$) = 0

You would have to set the limiting value at infinity be the value of V for the conducting planes if the planes are not at zero potential. Now, imposing this boundary condition will be severe. Your solution for V will then be pretty trivial.

Nevertheless, if the conducting planes are truly infinite in extent, then I don't see any other reasonable boundary condition! I believe the intent of the question was probably to determine the behavior of the potential near the origin for "large" plates that don't necessarily extend to infinity and for which no boundary conditions are specified for large r.

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TSny said:
You would have to set the limiting value at infinity be the value of V for the conducting planes if the planes are not a zero potential. Now, imposing this boundary condition will be severe. Your solution for V will then be pretty trivial.

Nevertheless, if the conducting planes are truly infinite in extent, then I don't see any other reasonable boundary condition! I believe the intent of the question was probably to determine the behavior of the potential near the origin for "large" plates that don't necessarily extend to infinity and for which no boundary conditions are specified for large r.

Or maybe,

lim $_{r→ -∞ }$V(r,θ) = 0

carloscariell said:
Or maybe,

lim $_{r→ -∞ }$V(r,θ) = 0

No, r is non-negative by definition.

Yes. You are right.

Guess that what I first meant to say is that

lim r→−∞ V(r, α/2 ± ε) = 0 , ε ≥ 0

In some area around the two planes where r→∞, V→0;

The other option would be V→∞ when r→∞. I don't think so, but I can't strongly argue the reason.

## 1. What is the Laplace equation on two semi-infinite conductor planes?

The Laplace equation on two semi-infinite conductor planes is a mathematical equation that describes the distribution of electric potential between two semi-infinite conducting planes. It takes into account the boundary conditions of the system and allows for the determination of the electric potential at any point between the two planes.

## 2. What are the boundary conditions for the Laplace equation on two semi-infinite conductor planes?

The boundary conditions for this equation are that the potential is constant on each of the two planes and that there is no potential gradient in the direction perpendicular to the planes.

## 3. How is the Laplace equation on two semi-infinite conductor planes solved?

The solution to this equation involves using separation of variables techniques and solving for the coefficients using the boundary conditions. This results in a series of sine and cosine terms that can be used to determine the electric potential at any point between the two planes.

## 4. What are the applications of the Laplace equation on two semi-infinite conductor planes?

This equation is commonly used in electrostatics and can be applied to problems involving parallel plate capacitors, electric field mapping, and the behavior of conductors in an electric field. It is also used in the design and analysis of electronic devices.

## 5. What are the limitations of the Laplace equation on two semi-infinite conductor planes?

This equation assumes that the two planes are infinite in size and that there are no other conductors or charges present in the system. It also does not take into account any time-dependent effects, such as charging or discharging of the conductors. Additionally, the solution may not be accurate in regions close to the edges of the planes.

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