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srn
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Homework Statement
A conducting sphere of radius R2 has a central cavity of radius R1 that holds a charge q in its centre. Determine the electrical field for r > R2, r < R1 and R1 < r < R2 and determine the charge density induced by q.
I'm not allowed to include a link to my figure, but I'm sure you understand. In 2D it's 2 circles with a common centre, one has radius R2 and the other R1 where R1 < R2.
Homework Equations
(not sure why it's not displaying LaTeX from this point on)[itex]∫e.dA = \frac{q}{\epsilon_0}[/itex]
The Attempt at a Solution
Q induces a charge on the sphere.
r < R2
The electrical field due to the sphere is zero (E-field inside hollow conductors in electrostatic equillibrium is zero). That means the E-field is only due the q.
For a random r < R2:
[itex]E = \frac{q}{\epsilon_0}\cdot 4\pi r^2[/itex]
R1 < r < R2
[itex]E \cdot 4\pi r^2 = \frac{q + (3/4 \pi r^3 - R_1^3)\rho}{\epsilon_0}[/itex]
r > R2
[itex]E \cdot 4\pi r^2 = \frac{q + (3/4 \pi R_2^3 - R_1^3)\rho}{\epsilon_0}[/itex]
I think this is correct, however, I still need the charge that is incuded on the sphere. Intuitively I would say it is equal to q itself, because the electrical field inside (the solid sections) of the sphere should be zero. Though I'm confused by the 1/r^2 decay; for a uniform field it's easy...
Come to think of it, suppose you start off with the sphere and somehow turn the charge off. Suppose you then turn it on, the electrons would then migrate torwards the cavity. While that happens there is an electrical field inside the sphere, but after a while these electrons should settle, more specifically when the e-field generated by the separation of e- and positive molecules counteracts the e-field generated by the charge. Right? So that would mean the e-field for r1 < r < r2 is zero?
second edit: then again, the field in r1 < r < r2 is opposite to that of the charge, so my solution for r > r2 cannot be correct as I add both. Do I need to substract the "weakening" of q's field as it passes through the sphere?
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