From c oulomb field of a point c harge to Gauss Law
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From coulomb field of a point charge to Gauss LawLecture 11We first define Electric Flux:
E-fieldlines
Consider E-lines hitting an area ΔA.The electric flux through ΔA is defined by:
For a point charge Q, the total electric flux hitting the spheres surface with radius r is given by:
In other words: Total Flux emitted by he point charge
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General Statement of Gauss Law
Define Gaussian surface S to be the surface which enclosed the charge
then flux emitted through S equates to: .
Turns out, S can be any arbitrary shape, Q can be any charge
distribution within S.
We have .
Porcupine-needle analogy:by counting the needles, one can determine the size of the porcupine.
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Field due to charges which have a spherical symmetry.
Find EP
So far as it is spherically symmetric:
Proof:through P draw Gaussian surface S
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Clicker 11-1
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Cylindrical SymmetryGiven: Long uniformly charged rod
Find: Field at P
P
Gaussian surface – cylinder through P
E? radially outwards
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Plane symmetry: one planeFrom analytic integration, results we have
ΔQ: in the shaded areaΔA: Ends of the cylinder
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Clicker 10-3