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If Er is constant, the potential 0 will be linear in r with the assumed form
0(r) = A(r - B) + C,
where A, B, and C are constants to be determined. If 0 is the potential at the inner
cylinder, r = r, then
#(r) = A(r - ra) + Oo.
Using (2.10) in (2.7a) and defining p(r) = po, then
aPO (r - ra) + 0,
Now a suitable equation for p(r) is found by inspection after substituting (2.Ila) into
(2.8). The charge density is then given by
p(r) = rapo
The boundary conditions on the outer cylinder, r = rb, have not yet been considered.
The assumption that E, is constant over the domain precludes specification of another
boundary condition on the outer cylinder.
A numerical boundary value problem
requires specification of an additional boundary condition to ensure that the solution
is unique. Requiring
(rb)= 0 determines the geometry of the outer cylinder. Using
the boundary condition in (2.11a) and solving for rb yields
rb = O60+
The analytical solution and geometry of the outer cylinder are fully defined if po, ra,
#o are