disc_3_cap_solutions - Physics 2524, Discussion 3 Solutions...

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Physics 2524, Discussion 3 Solutions 9/6/2011 Consider a “Capacitor” made of two thin, flat disks of radius R. The disks are perpendicular to the z-axis as shown, the first at z = 0 with charge Q 1 , the second at z = a with charge Q 2 . Assume the charge is distributed uniformly on the disks. Your job is to calculate the electric field along the z-axis. Making the definition of the surface charge density, σ = Q/A = Q/( π R 2 ) for a disk, the magnitude of the electric field due to one of the disks on the center (z) axis a distance d from the disk is A) Assume that Q 1 = Q and Q 2 = -Q. For the points on the z-axis: z = -a/4, z = a/2, and z = a/4, what are the directions of: the field due to the bottom disk, E 1 ; the field due to the top disk, E 2 ; and the total electric field due to both plates, E T ? z = -a/4: E 1 is down, E 2 is up z = a/4 and a/2: E 1 is up and E 2 is up B) If Q 1 = -Q 2 = 0.50 μ C (5.0x10 -7 C), R = 50 cm, and a = 4 cm, what is the electric field due to just the bottom disk for the points on the z-axis at z = -a/4, z = a/4, and z = a/2? σ
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disc_3_cap_solutions - Physics 2524, Discussion 3 Solutions...

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