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Unformatted text preview: Practice Midterm 2 Solution 1 Diamond Circuit a) First let’s calculate electric field due to two cylinder without dielectric. Uni formly charge inner cylinder with total charge + Q and uniformly charge outer cylinder with total charge Q . By symmetry around ˆ zaxis, we know electric field will be pointing in ˆ r direction. That is, vector E = E r ˆ r . Drawing gaussian pillvox around cylinder with radius r , we can use gauss’s law, contintegraldisplay vector E · vector A = Q enc ǫ (1) integraldisplay L integraldisplay 2 π E r ˆ r · ˆ rrdφdz = + Q ǫ (2) integraldisplay L integraldisplay 2 π E r rdφdz = Q ǫ (3) 2 πLE r r = Q ǫ (4) E r = Q 2 πǫ Lr (5) vector E = Q 2 πǫ Lr ˆ r (6) We can find potential difference between two cylinder by using equation Δ V = integraltext vector E · d vector l . Δ V = integraldisplay vector E · d vector l (7) V b V a = integraldisplay b a Q 2 πǫ Lr ˆ r · ˆ rdr (8) V b V a = integraldisplay b a Q 2 πǫ Lr dr (9) V b V a = Q 2 πǫ L ln a b (10) Since we are interested in potential drop V a V b = V , V = Q 2 πǫ L ln b a (11) 1 Definition of capacitance is C = Q V thus, C = 2 πǫ L ln b a (12) (You should have above expression in your equation sheet). But since there is dieletric between two cylinder C = KC = K 2 πǫ L ln b a (13) b) How to simplify capacitor is shown in Figure 1. Capacitors in series can be simplified by using relationship...
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This note was uploaded on 02/04/2010 for the course PHYSICS 7B taught by Professor Packard during the Spring '08 term at Berkeley.
 Spring '08
 Packard
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