Question

4+4: Physics 2524, Discussion Disc. Sec: 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 Qi, the second at z-a with charge Qr Assume the charge is distributed uniformly on the disks. Your job is to calculate the electric field along the z-axis. Disk at z = a, Radius R Charge Q Disk at z=0, Radius R Charge Q Making the definition of the surface charge density, σ-QA-0(1 R) 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 (G =8.85x10-12C2/Nm3 A) Assume that Qi Q and Q--Q. For four points on the z-axis: z--a/4, z-a/4, z-a/2 and z-5/4 a, what are the directions of: the field due to the bottom disk, Ei; the field due to the top disk, Es; and the total electric field due to both plates, Ef? B) IfQi =-Qs-0.50 pC (5.0x10, 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 -a/4, z-4, and C) For the same values for Q, R, and a as in part B, what is the total electric field, due to both disks for the points on the z-axis at z-a/4, z- a/4, and z a/2? D) A common approximation for the field close to charged plates is that the field is a constant equal to the electric field at the surface of the plate. How good of an approximation would this be for the case you calculated exactly in part C? What is the relative error (E-Espro/Esael for the three points?

Answer 1

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E_r represent total electric field due to both plates E_1 represent electric field due to Q_1 E_2 represent electric field due to Q_2 Q_1 = -Q_2 = Q at Z = -a/4 Direction of E_1 = -ve Z-axis Direction of E_2 = +ve Z-axis Direction of E_1 = -ve Z-axis (as E_1 > E_2) here as shown in fig 2 at Z = a/4 Direction of E_1 = +ve z-axis Direction of E_2 = +ve Z-axis Direction of E_T = +ve Z-axis at Z = a/2 Direction of E_1 +ve Z-axis Direction of E_2 = +ve Z-axis Direction of E_T = +ve Z-axis at Z = 5a/4 Direction E_1 = +ve Z-axis Direction of E_2 = -ve Z-axis Direction of E_T = -ve Z-axis \r\n Q_1 = 0.50 mu c R = 50 cm a = 4 cm Electric field due to disk at its central axis E = sigma/2 elementof_0 (1 - d/squareroot d^2 + R^2) d rightarrow distance from center on axis sigma