Electrostatic Field in a Long Metal Cylinder
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A long metal cylinder with radius a is coaxial with, and entirely inside, an equally long metal tube with internal radius 2a and external radius 3a. The space between the cylinder and the tube is filled with an LIH dielectric material with relative permittivity a. This space is also permeated by a uniform free charge distribution with density pf. outside the tube, the charge density is zero. (a) Write down Poisson's equation for the electrostatic potential in the dielectric material in the region between the cylinder and the tube. You should justify the choice of the coordinate system that you use. (b) Find the general solution of Poisson's equation in the dielectric material. Now assume that the metal cylinder and tube are both held at zero potential. (c) Using appropriate boundary conditions, show that the potential within the dielectric at distance r from the axis of symmetry is given by LlFa? (1 3 In 2 1n(r/a)) r21 V (r) = -Pf for a < 7' < 2a. oes )
(d) Determine the electrostatic field E in the dielectric material. (e) Obtain an expression for the free charge per unit length, A, on the inner cylinder.
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Solution Summary
The electrostatic fields in a long metal cylinder is examined.
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Poisson equation in a medium with relative permittivity is:
(1.1)
Since the system has cylindrical symmetry and can be considered infinite in the z-direction, we can conclude that the potential is independent of the z-coordinate and the equation becomes a two dimensional Poisson equation in polar coordinates:
(1.2)
To solve this equation we start with the associated Laplace equation:
(1.3)
And we assume that we can write the solution as a product of two univariate functions:
(1.4)
When we plug this back into equation (1.3) we see that the partial derivatives become full derivatives:
(1.5)
Dividing by we get:
(1.6)
Or:
(1.7)
Both sides are independent of each other (they each depend on a different variable), and since this must be true for any both sides must be equal the same constant:
(1.8)
We now look at the angular equation and we identify three cases. Furthermore, we require that the solution will be periodic, since a full revolution returns us to the same point and the potential must be a single value function.
Case 1:
The equation becomes:
(1.9)
The solution to this equation is:
(1.10)
However the periodic condition can be satisfied if and only if A=B=0.
Case 2:
The equation becomes:
(1.11)
The solution is:
(1.12)
Again, the ...
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