Discussion Session for PHY2049 Physics with Calculus 2 - Electromagnetism Spring 2012 Week # 12 Problem 1 In the figure, a parallel plate capacitor has square plates of edge length L= 1.0 m. A current ~ between the plates, of 2.0 A charges the capacitor, producing a uniform electric field E ~ perpendicular to the plates. (a) What is the displacement current id through the with E region between the plates? (b)What is dE/dt in this region? (c) What is the displacement H ~ around ~ · ds current encircled by the square dashed path of edge d=0.5m? (d) What is B this square dashed path? Solution a) The charge in a capacitor is related to the voltage across its plates by q = CV . In a parallel plate capacitor we have that C = 0 A s where s is the separation between the plates. Also, since the electric field inside the plates is almost uniform we have that V = Es, hence q = CV = CEs = 0 A Es = 0 AE s Taking the time derivative of the expression above we obtain dq dE = 0 A dt dt dE I = 0 A dt The displacement current is then id = 0 dΦE d(E · A) dE I = 0 = 0 A = 0 A =I dt dt dt 0 A Thus, the lesson is that the displacement current is ALWAYS equal to the current that is charging the capacitor. The “always” here means “at all times”. Therefore, for a current I = 2A the displacement current is id = 2A. b) We’ve already done the work for this question. This is dE I 2 = = = 2.3 × 1011 V/m·s dt 0 A 8.85 × 10−12 (1)2 1 c) Since the ratio id /Area is constant1 we have idenc = d2 0.52 Aenc id = 2 id = 2 2 = 0.5A Atot L 1 (1) d) Since there’s no conduction current through the plate, Ampere-Maxwell law reads I ~ = µ0 i d ~ · ds B (2) Note that usually the right hand side is µ0 ienc + µ0 id , but here there’s no real current enclosed by the loop, i.e. µ0 ienc = 0. Thus I ~ = µ0 id = 4π × 10−7 × 0.5 = 6.3 × 10−7 T·m ~ · ds B 1 The electric field inside the capacitor is uniform 2 (3)