Boundary Value Problem for Electrostatic Potential in a Channel

In summary, electrostatic potential V (x,y) in the channel - \infty < x < \infty, 0 \leq y \leq a satisfies the Laplace equation. The potential on the wall y = a is greater than the potential on the wall y = 0, due to the wall being earthed. By seeking a solution of an appropriate form, find V (x,y) in the channel. When in doubt, try separation of variables.
  • #1
MathematicalPhysics
40
0
I need some help starting off on this question.

Electrostatic potential [tex]V (x,y)[/tex] in the channel [tex]- \infty < x < \infty, 0 \leq y \leq a[/tex] satisfies the Laplace Equation

[tex]\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2}= 0[/tex]

the wall [tex]y = 0[/tex] is earthed so that

[tex]V (x,0) = 0[/tex]

while the potential on the wall [tex]y = a[/tex]

[tex]V (x,a) = V_0 \cos{kx}[/tex] where [tex]V_0 , k[/tex] are positive constants.

By seeking a soln of an appropriate form, find [tex]V (x,y)[/tex] in the channel.
 
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  • #2
When in doubt, try separation of variables!
V(x,y)=F(x)G(y).
Inserting this into Laplace, yields, by rearrangement:
[tex]\frac{G''(y)}{G(y)}=-\frac{F''(x)}{F(x)}[/tex]
 
  • #3
Thanks, I was trying to get it in the form [tex]V (x,y) = F (a) V_0 \cos{kx}[/tex] then sub the values into the Laplace eqn. Is that going about it the wrong way?
 
  • #4
You DO mean (in my notation!)
[tex]V(x,y)=G(y)V_{0}\cos(kx)[/tex]?
If so, then it will work.
Note that the separation of variables method in this case implies:
[tex]G''(y)=k^{2}G(y)[/tex]
 
  • #5
Hmm I am not really keeping up with you here sorry. How can I find things out about F(x) and G(y) (in your notation) when we have just introduced them?

--------edit----------

oh yes sorry I am with you, then differentiate V (x,y) = G(y) V_0 cos (kx) to get the different bits to go in the Laplace eqn? ahh!
 
Last edited:
  • #6
Your lightbulb switched on in your edit, I see..:wink:
Hint:
You should be able to see that G(y)=ASinh(ky), where A is some constant, and Sinh() the hyperbolic sine function.
 
  • #7
Now I've got G''(y) - k^2 G (y) = 0, still on the right track yeh?
 
  • #8
Yes, you are.
 
  • #9
MathematicalPhysics said:
Now I've got G''(y) - k^2 G (y) = 0, still on the right track yeh?

From this how do I get to G(y)=ASinh(ky)?
 
  • #10
Note that:
[tex]\frac{d}{dy}ASinh(ky)=kACosh(ky),\frac{d^{2}}{dy^{2}}ASinh(ky)=k^{2}ASinh(ky)[/tex]
This shows that ASinh(ky) is a solution for G(y).
Similarly, you may show that BCosh(ky) is another solution for G(y).
Your general solution is therefore:
G(y)=ASinh(ky)+BCosh(ky)
Apply the boundary condition at y=0 to prove that B=0
 
  • #11
Thanks for being so patient. How can I show that ASinh(ky) is a soln for G(y)? I am used to looking at the differential eqn and substituting

G(y) = e^{ry} so G'(y) = re^{ry} and G''(y) = r^2. e^{ry}

Is there no such substitution to show that ASinh(ky) & BCosh(ky) are solns?
 
  • #12
MathematicalPhysics said:
Thanks for being so patient. How can I show that ASinh(ky) is a soln for G(y)? I am used to looking at the differential eqn and substituting

G(y) = e^{ry} so G'(y) = re^{ry} and G''(y) = r^2. e^{ry}

Is there no such substitution to show that ASinh(ky) & BCosh(ky) are solns?
Your approach is ABSOLUTELY correct!
Do you agree with the following result, using your method:
[tex]r=\pm{k}[/tex]??
 
  • #13
arildno said:
Your approach is ABSOLUTELY correct!
Do you agree with the following result, using your method:
[tex]r=\pm{k}[/tex]??


Yeh which gives gen soln G (y) = Ae^{ky} + Be^{-ky}

but the boundary condition at y=0 doesn't allow for B so we are left with

G(y) = Ae^{ky} ?
 
  • #14
A bit too fast there..
Let's write the general solution for G(y) as follows:
[tex]G(y)=K_{+}e^{ky}+K_{-}e^{-ky}[/tex]
where the K's are constants to be determined by boundary conditions.
Prior to that step, however, let's rewrite the general solution as:
[tex]G(y)=(K_{+}+K_{-})\frac{e^{ky}+e^{-ky}}{2}+(K_{+}-K_{-})\frac{e^{ky}-e^{-ky}}{2}[/tex]
1. We now set [tex]B=K_{+}+K_{-},A=K_{+}-K_{-}[/tex]
(Clearly, A and B are as arbitrary as the K's!)
2) We now recognize:
[tex]Cosh(ky)=\frac{e^{ky}+e^{-ky}}{2}[/tex]
[tex]Sinh(ky)=\frac{e^{ky}-e^{-ky}}{2}[/tex]
Or, we may rewrite G(y) as:
[tex]G(y)=ASinh(ky)+BCosh(ky)[/tex]
 
  • #15
right now I need to find

B: iv done this and basically its because cosh can never be zero which implies B must be zero.

A: at y=a V(x,a) = V_0 cos{kx}

therefore for y=a, G(y)=1

ASinh(ka) = 1 is that right?

hmm I am not sure about this.
 
  • #16
Yep, so your solution is:
[tex]V(x,y)=V_{0}\frac{Sinh(ky)\cos(kx)}{Sinh(ka)}[/tex]
 
  • #17
Thank you so much, I understand much more now than I did even yesterday!
 

What is a boundary value problem for electrostatic potential in a channel?

A boundary value problem for electrostatic potential in a channel is a mathematical model that describes the distribution of electric potential in a confined space, such as a channel or a tube. It involves solving a set of differential equations with specified boundary conditions to determine the potential at different points in the channel.

What are the applications of boundary value problem for electrostatic potential in a channel?

The boundary value problem for electrostatic potential in a channel has numerous applications in various fields such as electrical engineering, physics, and biology. Some examples include the design of microfluidic devices, analysis of ion channels in biological cells, and the development of electrostatic sensors and actuators.

What are the boundary conditions in a boundary value problem for electrostatic potential in a channel?

The boundary conditions in a boundary value problem for electrostatic potential in a channel refer to the specified values of potential at the edges or surfaces of the channel. These conditions can be either Dirichlet boundary conditions, where the potential is known at the boundaries, or Neumann boundary conditions, where the derivative of the potential is known at the boundaries.

How is the electrostatic potential in a channel calculated?

The electrostatic potential in a channel is calculated by solving a set of differential equations that describe the behavior of electric potential in the channel. These equations, known as the Poisson or Laplace equations, take into account the charge distribution and the boundary conditions to determine the potential at different points in the channel.

What are some numerical methods used to solve a boundary value problem for electrostatic potential in a channel?

There are several numerical methods used to solve a boundary value problem for electrostatic potential in a channel, including the finite difference method, finite element method, and boundary element method. These methods involve discretizing the channel into smaller elements and then solving the resulting equations using computer algorithms.

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