Electrostatics Boundary condition and coordinate choice

[Mike Jonese]

Homework Statement

So I have an equation V = Ae^(kx)+Be^(-kx)
And boundary conditons V= V₀ when x=0 and V= 0 when x=b

2. Homework Equations
I have solved ones where v=0 at x=0 where it nicely simplifies as the exponentials =1 and the Coeffecients A=-B which leads to a sinh function and I can handle that. But, for problems where V = V₀ right on x=0 I can't simplify it in a convenient way

The Attempt at a Solution

when x=b Ae^(kb) + Be^(-kb) = 0
I can't figure out a way to simplify this expression to figure out the coeffecients.
This is a 3 dimensional problem but I am just focusing on the x aspect right now. I can imagine shifting the coordinate frame to conveniently place V=0 at x=0 but I really would like to avoid doing that and figure out how to just solve this at with the given coordinate frame.
Thanks so much for any insight

 

[Charles Link]

What is wrong with $A+B=V_o$ and $0=Ae^{kb}+Be^{-kb}$? It's still basically two equations and two unknowns.

 

[Mike Jonese]

What is wrong with $A+B=V_o$ and $0=Ae^{kb}+Be^{-kb}$? It's still basically two equations and two unknowns.

Well, for more background, the entire problem is a grounded box with each side of length a, the only side with Constant Potential V0 is insulated from the others, lies in the yz plane and extends from z=0 to z=a and y=0 to y=a. That potential is on that face at x=0.
BC.s V=0 when y=0, V=0 when y=a
V=0 when z=0, V=0 when z=a
V=V0 when x=0, V=0 when x=a
My potential fxn at this point is V(x,y,z)= (Ae^{√(k^2+l^2)x} + Be^{-√(k^2+l^2)x} )(Csin(ky))(Dsin(lz))
I will have to sum these equations up and find a constant to weight them with. So, I wanted to find a way to consolidate A and B, so that i can absorb all the constants together in C_n before I sum everything. This was easier when the potential was not directly on x=0 as described above. Is this more clear?

 

[Charles Link]

What you have is rather complex. I don't know that the assumptions you made about the potential function are even correct without seeing more additional detail. In general, unless the geometries are very simple, electrostatic problems with boundary values are very much non-trivial.

 

[Mike Jonese]

What you have is rather complex. I don't know that the assumptions you made about the potential function are even correct without seeing more additional detail.

Ok. The potential is constant. I used laplace eqn to say the ∇²V(x,y,z)= 0
Im guessing function is something like V(x,y,z)= X(x)Y(y)Z(z)
I used separation of variables to get
(1/X)∂²X/(∂x^2) + (1/Y)∂²Y/(∂y^2) + (1/Z)∂²Z/(∂z^2) = 0
This implies that each term is a constant.
C1 + C2 + C3 =0 Potential cycles between two points where it is zero in Y and in Z = this implies sin,cos solution and negative constants C2,C3
Potential is a some constant value and then eventually zero along x, this implies C1 is positive
C1 -C2 -C3 = 0
C1 = C2 + C3 let C2 = k^2 let C3 = l^2
(k^2+l^2) = k^2 + l^2
X(x)= Ae√(k^2+l^2)x + Be-√(k^2+l^2)x
Y(y)= Csin(ky)+Dcos(ky)
Z(z)= Esin(lz)+Fcos(lz)
Plugging all that into my "guess Function"
V(x,y,z)= (Ae√(k^2+l^2)x + Be-√(k^2+l^2)x )(Csin(ky)+Dcos(ky))(Esin(lz)+Fcos(lz))
The Boundary conditions eliminate the coefficients D and F, and cosines, to allow v=0 at y=0 and z=0
Leaves us with
V(x,y,z)= (Ae√(k^2+l^2)x + Be-√(k^2+l^2)x )(Csin(ky))(Dsin(lz)) where I just changed constant E to a new Constant D
That is how I got to this point. In my very first post I just "k" as the constant in the exponential to simplify things.

 

[Charles Link]

Ok. The potential is constant. I used laplace eqn to say the ∇²V(x,y,z)= 0
Im guessing function is something like V(x,y,z)= X(x)Y(y)Z(z)
I used separation of variables to get
(1/X)∂²X/(∂x^2) + (1/Y)∂²Y/(∂y^2) + (1/Z)∂²Z/(∂z^2) = 0
This implies that each term is a constant.
C1 + C2 + C3 =0 Potential cycles between two points where it is zero in Y and in Z = this implies sin,cos solution and negative constants C2,C3
Potential is a some constant value and then eventually zero along x, this implies C1 is positive
C1 -C2 -C3 = 0
C1 = C2 + C3 let C2 = k^2 let C3 = l^2
(k^2+l^2) = k^2 + l^2
X(x)= Ae√(k^2+l^2)x + Be-√(k^2+l^2)x
Y(y)= Csin(ky)+Dcos(ky)
Z(z)= Esin(lz)+Fcos(lz)
Plugging all that into my "guess Function"
V(x,y,z)= (Ae√(k^2+l^2)x + Be-√(k^2+l^2)x )(Csin(ky)+Dcos(ky))(Esin(lz)+Fcos(lz))
The Boundary conditions eliminate the coefficients D and F, and cosines, to allow v=0 at y=0 and z=0
Leaves us with
V(x,y,z)= (Ae√(k^2+l^2)x + Be-√(k^2+l^2)x )(Csin(ky))(Dsin(lz)) where I just changed constant E to a new Constant D
That is how I got to this point. In my very first post I just "k" as the constant in the exponential to simplify things.

I think the assumption that $V(x,y,z)=V_1(x)V_2(y)V_3(z)$ is quite a huge assumption. The problem doesn't have spherical symmetry, but if it did this assumption would not work. Perhaps it is justifiable, but I would need much further study to determine whether it indeed is applicable. Assuming each term is a constant, I don't agree with your solutions. The solutions would be of the form $V(x)=Ax^2+Bx+C$, etc.

 

[Mike Jonese]

I think the assumption that $V(x,y,z)=V_1(x)V_2(y)V_3(z)$ is quite a huge assumption. The problem doesn't have spherical symmetry, but if it did this assumption would not work. Perhaps it is justifiable, but I would need much further study to determine whether it indeed is applicable. Assuming each term is a constant, I don't agree with your solutions. The solutions would be of the form $V(x)=Ax^2+Bx+C$, etc.

I'm getting the technique from Griffiths electrodynamics E&M in Chapter 3. No problem I appreciate you taking a look at this

 

[TSny]

V(x,y,z)= (Ae√(k^2+l^2)x + Be-√(k^2+l^2)x )(Csin(ky))(Dsin(lz)) where I just changed constant E to a new Constant D
That is how I got to this point. In my very first post I just "k" as the constant in the exponential to simplify things.

I think you're OK up to here. You should be able to use boundary conditions at y = a and z = a to determine the allowed values for k and l.

If you factor A out of (Ae√(k^2+l^2)x + Be-√(k^2+l^2)x ), you can then lump all the constants A, C, and D, into one constant. The ratio B/A which appears when you factor A out can be determined from the boundary condition on X(x) at x = b.

I'm a little confused with what b represents. You mentioned that you have a box of side length a. So, shouldn't the boundary of x be at x = a? Why use the symbol b? I'm referring here to your original post where you wrote:

And boundary conditons V= V₀ when x=0 and V= 0 when x=b

 

[Charles Link]

I missed the factor $1/V_x$ in your equation $\frac{1}{V_x} \frac{\partial^2 V_x}{\partial x^2}=Constant$. Ignore my comment that the solution needs to be of the form $V=Ax^2+Bx+C$.

 

[Mike Jonese]

I think you're OK up to here. You should be able to use boundary conditions at y = a and z = a to determine the allowed values for k and l.

If you factor A out of (Ae√(k^2+l^2)x + Be-√(k^2+l^2)x ), you can then lump all the constants A, C, and D, into one constant. The ratio B/A which appears when you factor A out can be determined from the boundary condition on X(x) at x = b.

I'm a little confused with what b represents. You mentioned that you have a box of side length a. So, shouldn't the boundary of x be at x = a? Why use the symbol b? I'm referring here to your original post where you wrote:

Oh sorry that was an error, I didn't know how much of the problem would be necessary to include. That first post should have been when x=a.
So I tried what you recommended, this is what I got.
V(x,y,z) = (Ae^{√(k^2+l^2)x}+Be^{-√(k^2+l^2)x})(Csin(ky))(Dsin(lz)
V(x,y,z) =A (e^{√(k^2+l^2)x}+(A/B)e^{-√(k^2+l^2)x})(Csin(ky))(Dsin(lz)
V(x,y,z) =C (e^{√(k^2+l^2)x}+(A/B)e^{-√(k^2+l^2)x})(sin(ky))(sin(lz)) absorbed into C (calling this "eqn 1.1")
then to find (A/B)
at x=a
(e^{√(k^2+l^2)a}+(A/B)e^{-√(k^2+l^2)a}) = 0
(B/A) = (-e^{√(k^2+l^2)a}) / ( e^{-√(k^2+l^2)a}))
(B/A) = -e^{2√(k^2+l^2)a} so that's my constant,
Plugging back into "egn 1.1"
V(x,y,z) =C (e^{√(k^2+l^2)x}+(-e^{2√(k^2+l^2)a})e^{-√(k^2+l^2)x})(sin(ky))(sin(lz))
I still kind of have the same problem with the exponential constant (B/A) part in the first parentheses. I did get the constant off the very first exponential term. Am I missing something obvious?

 

[Mike Jonese]

Actually I guess at this point would I be able to say I have
V(x,y,z) =C (e^{√(k^2+l^2)x}-e^{2a(k^2+l^2)x})(sin(ky))(sin(lz))
and then just use the boundary conditions V=0 at y=a to get k=(n*pi/a) and V=0 at z=a to get l=(m*pi/a) and use fouriers prthagonal trick and sum to find the constant C?

 

[TSny]

Actually I guess at this point would I be able to say I have
V(x,y,z) =C (e^{√(k^2+l^2)x}-e^{2a(k^2+l^2)x})(sin(ky))(sin(lz))
and then just use the boundary conditions V=0 at y=a to get k=(n*pi/a) and V=0 at z=a to get l=(m*pi/a) and use fouriers prthagonal trick and sum to find the constant C?

Yes, that should work.

Your exponent for the second exponential appears to have a typographical error. It should be dimensionless.

 

[Mike Jonese]

Yes, that should work.

Your exponent for the second exponential appears to have a typographical error. It should be dimensionless.

Hmmm on that one I multiplied (B/A) into my exponential
(-e^{2√(k^2+l^2)a})e^{-√(k^2+l^2)x})
That is where my x is coming from. Is that what you meant?

 

[TSny]

Hmmm on that one I multiplied (B/A) into my exponential
(-e^{2√(k^2+l^2)a})e^{-√(k^2+l^2)x})

You have the product of two exponentials, so the exponents add. But this doesn't yield what you wrote in #11.

 

[Mike Jonese]

Thanks for all the help!