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Why do we need to pick a very small A for curved surfaces in electrostatics?
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[QUOTE="Tsunoyukami, post: 4498562, member: 274774"] Ah, my apologies. I thought you had simply typed the inequalities incorrectly. As I stated, I would refer to section 2.3.5 Summary: Electrostatic Boundary Conditions on page 87. The section essentially says that the electric field is discontinuous at boundaries. I believe that for your purposes solving the problem under the conditions r < a, a < r < b, and r < b should suffice. I think its reasonable to assume that when r = a the "enclosed" charge is zero because when r = a there is no charge [I]inside[/I] this surface (it is instead an infinitesimal amount of charge smeared over the surface of the sphere described by ##V = \frac{4}{3} \pi r^{3}## which is not contained [I]within[/I]). Furthermore, it only makes sense to me that the field in each of the regions should have a formula valid for the entire region - that is, for r < a we know E = 0; if the problems asks us to find the field for r <= a I would expect to have a single expression valid in that entire region, including when r = a, namely E = 0. However, if in doubt I would ask your professor as to what he considers the proper interpretation to be. [/QUOTE]
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Introductory Physics Homework Help
Why do we need to pick a very small A for curved surfaces in electrostatics?
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