When one does an integral over a flat disk of charge, the result is that the electric field above the center of the disk is: (see Open Stax Example 5.8 and Young & Freedman Example 21.11) E(a) = 260 Vac2 + R2 In this expression: R is the radius of the disk " = Q/ A is the charge per area on the disk (A = TR2) x is the distance from the center of the disk (perpendicular to the disk) Co = 8.85 x 10-12 C2 / (Nm2) as defined. For a positive charge, the field points away from the disk. Considering this result for the electric field: . The electric field has a finite value at the surface, x = 0, unlike the fields due to point and line charges. . The magnitude of the electric field decreases for points away from the disk, in particular for x > O. . The magnitude of the electric field goes to zero very far from the disk, x - co Consider a charged disk with: R = 4.53 cm ( 1 cm = 10-2 m) Q = 7.18 uC (1 MC = 10-6 C) Define E(O) as the magnitude of the electric field at the surface of the disk. For what distance, x, will the electric field have the magnitude: E(x) = 0.62 E(0) In other words, at what distance from the disk will the field be a factor of 0.62 smaller than the field at the surface of the plate