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fluidistic
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Homework Statement
Two concentric conducting spheres of inner and outer radii a and b, respectively, carry charges +/-Q. The empty space between the spheres is half-filled by a hemispherical shell of dielectric (of dielectric constant ##\varepsilon / \varepsilon _0## ).
a)Find the electric field between the spheres.
b)Calculate the surface-charge density distribution on the inner sphere.
c)Calculate the polarization-charge density induced on the surface of the dielectric at r=a.
Homework Equations
Laplace equation, ##\triangle \Phi =0## for the region between the 2 spheres.
The Attempt at a Solution
Working on a).
To get ##\vec E## between the 2 spheres I'll go for the potential first and then use the relation ##\vec E =- \vec \nabla \Phi##.
I use spherical coordinates ##(r, \theta, \phi)##. I choose my z-axis such that the half-filled (with the dielectric) hemisphere is for ##z \leq 0##. This way I get that the potential has no dependency on the coordinate phi. In other words I get azimuthal symmetry which simplifies greatly the expression of the solution to Laplace equation.
Solving the Laplace equation in spherical coordinates leads to ##\Phi (r, \theta ) = \sum _{l=0}^ \infty [A_l r^l + B_l r^{-(l+1)}] P_l (\cos \theta )## (eq. 3.33 in Jackson's book).
Now I'm looking for the boundary conditions. The fact that the 2 spheres are conductors means that the potential over both surfaces is constant. I'll call ##V_a## the potential ##\Phi (r=a)## and ##V_b## the potential ##\Phi (r=b)##.
The boundary conditions are then ##\begin{array} \Phi (a, \theta ) = \sum _{l=0}^ \infty [A_l a^l + B_l a^{-(l+1)}] P_l (\cos \theta ) = V_a \\ \Phi (b, \theta ) = \sum _{l=0}^ \infty [A_l b^l + B_l b^{-(l+1)}] P_l (\cos \theta ) =V_b \end{array} ##.
(I've no idea why the latex doesn't "work" here, it should read Phi(a, theta) at the start of the first line).
I notice that ##V_a = V_a P_0 (\cos \theta )##. So that the infinite series collapses to only 1 term, the one with ##l=0##.
This yields, if I made no mistake, ##B_0=\frac{V_a-V_b}{\left ( \frac{1}{a} - \frac{1}{b} \right ) }## and ##A_0 = V_a -\frac{(V_a-V_b)}{\left ( 1 - \frac{a}{b} \right ) }##.
This is where I'm stuck. Previously, I was always given the potential on the surface of the conductor(s). But here I have ##V_a## and ##V_b## which are also unknowns. I must get them in terms of Q, a and b I guess. I just don't know how to proceed.
I've been thinking about it and I don't even have the slighest idea on how to get those constant potentials. Any idea would be welcome.