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Prof. D. Wilton ECE Dept. Notes 12 ECE 2317 Applied Electricity and Magnetism Notes prepared by the EM Group, University of Houston
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Example y x x = - d/2 x = d/2 (a) x > d/2, (c) x < -d/2, (b) -d/2 < x <d/2, By superposition,
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Example (cont.) y x x = - d/2 x = d/2 (a) x > d/2, (c) x < - d/2, (b) - d/2 < x < d/2, Choose:
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Example (cont.) -d/2 < x < d/2 ideal parallel-plate capacitor y x h
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Example (cont.) y x x = - d/2 x = d/2 (a) x > d/2, (c) x < - d/2, (b) -d/2 < x < d/2, This time, choose:
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Example (cont.) y x h ideal conductor of thickness h
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Perfect Electric Conductors (PEC) [S/m] A B PEC: Also, since (Ohm’s law)
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Perfect Electric Conductors (PEC) Inside PEC: Proof: assume point where v > 0 Only s on the surface is allowed S assume v > 0 inside small volume Contradiction ! + + + + + + +
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Perfect Electric Conductors (PEC) Inside an empty, hollow PEC shell there is no electric field Only s on the outer surface is allowed PEC shell + + + + + + + + + + + + E = 0 + “Faraday-cage effect”:
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Perfect Electric Conductors (PEC) Proof: Assume an electric field exists inside the shell at some point PEC shell + + + + + + + + + + + + Assume E 0 A B A flux line must exist through this point. The flux line must end on the shell (otherwise it contradicts Gauss’s law) V AB 0 Contradiction ! + There is also no charge density on the inner surface (otherwise, there would be a flux line coming from the inner surface). Hence, there is no electric field (and no flux lines) inside the cavity.
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Perfect Electric Conductors (cont.) Faraday-cage effect The men in the Faraday cages are protected from the high voltage arcs of the Tesla coil!
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Perfect Electric Conductors (cont.) Faraday-cage effect Entrance to a Faraday room Faraday shield at Art Nouveau power plant in Heimbach, Germany
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Shielding and Grounding q Spherical PEC shell Neutral (no charge) a b Drill hole and insert point charge, then solder hole. Neutral shell (no net charge) Find E : q a b
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(a) r < a (c) r > b (b) a < r < b q a b Neutral shell (PEC) Shielding and Grounding
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q (outside metal) The neutral metal shell does not block the static electric field
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Shielding and Grounding Find Q A, Q B : q - - - - - - - - + + + + + + + + + + + + + + + + QAQA QBQB q QAQA QBQB Neutral shell
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Shielding and Grounding q - - - - - - - - + + + + + + + + + + + + + + + + QAQA QBQB (since neutral shell) S so Hence and A Gaussian surface is chosen inside the metal shell.
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Example (cont.) Alternative solution (Gauss’s law): y x h PEC S + + + + + + + + A (The electric field outside the slab is the same as from a single sheet of surface charge.)
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Next, “ground” the shell: q - - - - - - - - -q E = 0 PEC wire Earth r > b : E = 0 A B This must hold for every path. proof: The earth is modeled as a very large conductor. If the electric field were non-zero at a point, there would be a flux line through the point ending on conductors and implying a voltage drop between them flux line
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charge on outer surface: q - - - - - - - - QBGQBG -q PEC wire Hence r > b: E = 0 Earth
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charge on outer surface (cont.): The charge q on the outer surface flows down to ground. q - - - - - - - - Q B G = 0 -q + + + + PEC wire Earth q - q q q before grounding after grounding
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Example Given: y x h r PEC Find E everywhere (This is the total surface charge density per square meter on the metal slab, which is the sum of the surface charge densities on the upper and lower surfaces.)
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Example Charge Model: y x h r PEC y x x = - h/2 x = h/2 + + + + + + + + + + + + + + + +
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Example y x x = - h/2 x = h/2 + + + + + + + + + + + + + + + + By superposition,
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Example y x h PEC The surfaces are initially charged as shown above. The bottom of the plate is then grounded. Find the charge and field everywhere
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Example (cont.) (a) charge on lower surface h PEC S A E = 0 ++++ ++++++++ ++++ Hence Earth
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Example (cont.) (b) charge on earth h PEC S ++++ ++++ ++++ ++++ A Hence Earth
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Example (cont.) (c) charge on top surface Hence So (After grounding) E = 0 (inside earth) x ++++++++++++++++++++
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Example (cont.) (before grounding) PEC (after grounding) PEC zero Earth
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