Solution to Laplace equation in 1D π2 π = 0, β π = ππ₯ + π 2 ππ₯ V Mean value theorem in 1D 1 π π₯ = [π π₯ + π + π π₯ β π ] 2 V(x-a) V(x) V(x+a) x-a x x+a x Mean value theorem in 2D π ππ V y ππ R x 1 = 2ππ π π ππ ππππππ q r z ο± Equipotentials R y x q Mean value theorem in 3D z 1 π π = 4ππ 2 r ο± ππ R y x πππ π πβπππ q1 q2 πππ£π 1 = 4ππ 2 π1 + π2 + π3 ππ π πβπππ r z ο± 1 = 4ππ 2 1 + 4ππ 2 ππ R y 1 + 4ππ 2 π1 ππ π πβπππ π2 ππ π πβπππ π3 ππ π πβπππ q3 = π1 π + π2 π + π3 π x =π π Mean value theorem π»2π = 0 1 π π = 4ππ 2 πππ π πβπππ No absolute maximum or minimum allowed in the region that π» 2 π = 0 There can never be a maximum or minimum in potential if π»2π = 0 Let there be two possible potentials which are solutions to Laplaceβs equation π V1 Same boundary conditions π V2 V3 = V1 - V2 Let there be two possible solutions: π Charge density π Q2 Q1 Q3 π π» β πΈ1 = π0 π π» β πΈ2 = π0 E = 0 inside the metallic regions Q3 Q2 π π’πππππ 1 π1 πΈ1 β ππ = π0 πΈ1 β ππ = π2 π0 πΈ1 β ππ = π3 π0 π π’πππππ 2 Q1 π π’πππππ 3 π π’πππππ π ππ πΈ1 β ππ = π0 πΈ1 β ππ = ππ’π‘ππ πππ’πππππ¦ ππ‘ππ‘ π0 E = 0 inside the metallic regions πΈ2 β π π = Q3 Q2 π π’πππππ 1 π π’πππππ 2 Q1 π π’πππππ 3 π2 πΈ2 β π π = π0 π3 πΈ2 β π π = π0 πΈ2 β π π = π π’πππππ π ππ π0 πΈ2 β π π = ππ’π‘ππ πππ’πππππ¦ π1 π0 ππ‘ππ‘ π0 Let πΈ3 β‘ πΈ1 β πΈ2 π» β πΈ3 = π» β πΈ1 β πΈ2 = 0 Q3 πΈ3 β ππ = 0 π π’πππππ 1 Q2 Q1 πΈ3 β ππ = 0 π π’πππππ 2 πΈ3 β ππ = 0 π π’πππππ 3 πΈ3 β π π = 0 π π’πππππ π πΈ3 β ππ = 0 ππ’π‘ππ πππ’πππππ¦ The corresponding potentials π1 , π2 , π3 are constants at the surfaces of the conductors. Q3 π» β π3 πΈ3 = π3 π» β πΈ3 + πΈ3 β π»π3 = 0 β πΈ32 Q2 Q1
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