Solving Poisson's Equation: Uniformly Charged Cylinder
Consider an infinitely long cylinder with uniform charge density and radius 'a'. Demonstrate Find the potential V both inside and outside the cylinder.
QUESTIONS
(Score is number right minus number wrong.)
From the symmetry of the problem, we can conclude that the potential V depends only on s (not on phi or z in cylindrical coordinates).
True
False
INSIDE the cylinder, the differential equation for the potential V is,
Equation (5a)
Equation (5b)
Equation (6c)
Equation (6d)
Equation (7e)
OUTSIDE the cylinder, the differential equation for the potential V is,
Equation (5a)
Equation (5b)
Equation (6c)
Equation (6d)
Equation (7e)
If we have the differential equations correctly, it is just a matter of integrating. So, take a moment to integrate these two equations (each twice) to get a solution V(s) for each.
INSIDE the cylinder, the solution for the potential V is,
Equation (8a)
Equation (8b)
Equation (8c)
Equation (8d)
Equation (8e)
Equation (8f)
OUTSIDE the cylinder, the solution for the potential V is,
Equation (8a)
Equation (8b)
Equation (8c)
Equation (8d)
Equation (8e)
Equation (8f)
Now we have solutions that actually satisfy the appropriate differential equations (Poisson and Laplace, respectively), but the solutions don't necessarily satisfy the boundary conditions. What are the boundary conditions?
A good place to arbitrarily set V=0 is for s at infinity.
True
False
It is permissible to set V=0 at s=a.
True
False
The only way that Equation (8f) can remain finite at s=0 is if
C1=0
C2=0
Both C1 and C2 equal 0
If V=0 at s=a in Equation (8f), we must have,
Equation (9a)
Equation (9b)
Equation (9c)
If V=0 at s=a in Equation (8e), we must have,
Equation (10a)
Equation (10b)
Equation (10c)
Equation (10d)
We must have V go to zero as s goes to infinity, so we must have D1=0.
True
False
The normal derivative of V (here dV/ds) is always continuous across a boundary.
True
False
The normal derivative of V is continuous across a boundary if there is no surface charge on the boundary because the normal component of E is continuous where no surface charge exists.
True
False
The correct value of the constant D1 is
Equation (11a)
Equation 11b)
Equation 11c)
Equation (11d)
The potential inside and outside the cylinder are correctly given by Equation (12).
True
False
EQUATIONS
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