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From c oulomb field of a point c harge to Gauss Law

Lecture 11. From c oulomb field of a point c harge to Gauss Law. We first define Electric Flux:. Consider E-lines hitting an area ΔA. The electric flux through ΔA is defined by: . E-field lines.

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From c oulomb field of a point c harge to Gauss Law

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  1. Lecture 11 From coulomb field of a point charge to Gauss Law We first define Electric Flux: Consider E-lines hitting an area ΔA. The electric flux through ΔA is defined by: E-field lines 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

  2. 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.

  3. 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

  4. Clicker 11-1

  5. Cylindrical Symmetry Given: Long uniformly charged rod P Gaussian surface – cylinder through P Find: Field at P E? radially outwards

  6. Plane symmetry: one plane From analytic integration, results we have ΔQ: in the shaded area ΔA: Ends of the cylinder

  7. Clicker 10-3

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