Prof. D. Wilton ECE Dept. Notes 12 ECE 2317 Applied Electricity and Magnetism Notes prepared by the EM Group, University of Houston

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,

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:

Example (cont.) -d/2 < x < d/2 ideal parallel-plate capacitor y x h

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:

Example (cont.) y x h ideal conductor of thickness h

Perfect Electric Conductors (PEC)  [S/m] A B PEC: Also, since (Ohm’s law)

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 !

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”:

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.

Perfect Electric Conductors (cont.) Faraday-cage effect The men in the Faraday cages are protected from the high voltage arcs of the Tesla coil!

Perfect Electric Conductors (cont.) Faraday-cage effect Entrance to a Faraday room Faraday shield at Art Nouveau power plant in Heimbach, Germany

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

(a) r < a (c) r > b (b) a < r < b q a b Neutral shell (PEC) Shielding and Grounding

q (outside metal) The neutral metal shell does not block the static electric field

Shielding and Grounding Find Q A, Q B : q QAQA QBQB q QAQA QBQB Neutral shell

Shielding and Grounding q QAQA QBQB (since neutral shell) S so Hence and A Gaussian surface is chosen inside the metal shell.

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.)

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

charge on outer surface: q QBGQBG -q PEC wire Hence r > b: E = 0 Earth

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

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.)

Example Charge Model: y x h r PEC y x x = - h/2 x = h/

Example y x x = - h/2 x = h/ By superposition,

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

Example (cont.) (a) charge on lower surface h PEC S A E = Hence Earth

Example (cont.) (b) charge on earth h PEC S A Hence Earth

Example (cont.) (c) charge on top surface Hence So (After grounding) E = 0 (inside earth) x

Example (cont.) (before grounding) PEC (after grounding) PEC zero Earth