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unscientific
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Homework Statement
Part (a): List the boundary conditions
Part (b): Show the relation for potential is:
Part (c): Find Potential everywhere.
Part (d): With a surface charge, where does the Electric field disappear?
Homework Equations
The Attempt at a Solution
Part (a)
Boundary conditions are:
[tex]E_{in}^{||} = E_{out}^{||}[/tex]
[tex]D_{out}^{\perp} - D_{in}^{\perp} = \frac{\sigma}{\epsilon_0}[/tex]
Part (c)
First we find potential inside and outside.
1. For potential inside ##V_{in}##, it is finite when r = 0, so ##b_n = 0##. When r = 0, ##V_{in} = 0##, so ##a_0 = b_0 = 0##.
2. For potential outside ##V_{out}##, as r → ∞, ##V \propto \frac{1}{r}## (From part (b)). So only term n = 1 exists.
3. For n=1, and ##\phi = \frac{\pi}{2}##, ##V_{in} = V_{out} = 0##, so sine term goes away: ##c_n = 0##.
Now we have the two potentials:
[tex]V_{in} = a_1 r cos(\phi)[/tex]
[tex]V_{out} = \frac{d_1}{r} cos(\phi)[/tex]
Applying first boundary condition at r = a,
[tex]\frac{\partial V_{in}}{\partial \theta} = \frac{\partial V_{out}}{\partial \theta}[/tex]
[tex] a_1 a^2 = d_1[/tex]
Applying second boundary condition at r = a,
[tex]D_{out}^{\perp} = D_{in}^{\perp} [/tex]
[tex]\epsilon_0 \frac{\partial V_{in}}{\partial r} = \frac{\partial V_{out}}{\partial r}[/tex]
[tex] \epsilon_0 a_1 = -\frac{\epsilon d_1}{a^2}[/tex]
Solving them, this implies ##\epsilon_0 + \epsilon = 0##, which doesn't make sense.