- #1
aftershock
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Homework Statement
Homework Equations
∇2V = 0
The Attempt at a Solution
I'm guessing I need to use Laplace's equation in 2 dimensions since the potential depends on x and y. I have no idea how I would do this.
vela said:This is a pretty standard type of problem. Consult your textbook for similar examples, and then post your work.
vela said:That's surprising. You could check another E&M text. It's also covered in math methods texts, like Arfken, as well. Look up "separation of variables" and "Laplace's equation".
aftershock said:By the 4th boundary condition:
Vo(y) = Asinh(n∏a/b)sin(n∏y/b)
solved for A, then plugged that into Asinh(n∏x/b)sin(n∏y/b)
I guess I can't do that since Vo(y) isn't constant? Still though... why?
Each value of n corresponds to a solution Vn, so you should writeaftershock said:Ok so I tried working out a solution but I went wrong somewhere.
From separation of variables I got:
V(x,y) = (Aekx+Be-kx)(Csin(ky)+Dcos(ky))
Boundary conditions:
V=0 when x=0
V=0 when y=0
V=0 when y=b
V=v0(y) when x=a
D=0 by the second condition, and I can then absorb C into A and B:
(Aekx+Be-kx)sin(ky)
-A = B by the first condition:
A[ekx-e-kx]sin(ky) = Asinh(kx)sin(ky)
k = n∏/b by the third condition:
Asinh(n∏x/b)sin(n∏y/b)
Now I'm pretty sure I'm supposed to do some sort of linear combination here to solve for A.
I did this instead. I'm pretty sure it's wrong but I'm not sure why?
By the 4th boundary condition:
Vo(y) = Asinh(n∏a/b)sin(n∏y/b)
solved for A, then plugged that into Asinh(n∏x/b)sin(n∏y/b)
I guess I can't do that since Vo(y) isn't constant? Still though... why?
Electric potential inside a pipe refers to the amount of electric potential energy per unit charge at a specific point inside the pipe. It is a measure of the electric field's strength at that point.
The electric potential inside a pipe is calculated using the formula V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the charge, and r is the distance from the center of the pipe.
The electric potential inside a pipe is affected by the charge of the particles inside the pipe, the distance from the center of the pipe, and the material properties of the pipe such as its conductivity and shape.
Yes, the electric potential inside a pipe can be negative. This indicates that the electric field is directed in the opposite direction of the electric charge.
The electric potential inside a pipe is related to electric current through Ohm's law, V = IR, where V is the electric potential, I is the current, and R is the resistance of the pipe. As the electric potential increases, the current also increases.