# Electric field due to a charged disc

A uniformly charged disk of radius 35.0 cm carries charge with a density of 7.90 x 10^-3 C/m^2. Calculate the electric field on the axis of the disk at

(a) 5.0 cm,

(b) 10.0 cm,

(c) 50 cm and

(d) 200 cm from the center of the disk.

https://brainmass.com/physics/electric-magnetic-fields/electric-field-due-charged-disc-125632

#### Solution Preview

** Please see the attached file for the complete solution response **

Fig. shows a uniformly charged disc (radius R = 0.35m), with charge density 7.9x10-3 C/m2. Let us consider an annular ring of the disc, co-centric with the disc, having radius r and width dr. Let us consider an infinitesimally small element of the ring between the radius vectors making angles Î¸ and Î¸+dÎ¸ with the X axis. Area of the element = rdÎ¸dr. Charge on the element = dq = Ïƒ rdÎ¸dr where Ïƒ is the charge density.

Electric field at point P on the axis of the disc, at a distance s from its centre, due to the element = dE = (1/4Î Ð„0)dq/l2 = (1/4Î Ð„0)Ïƒ rdÎ¸dr/l2

From the fig. l2 = s2+r2. Hence, dE = (1/4Î Ð„0)Ïƒ rdÎ¸dr/(s2+r2) .......(1)

The electric field vector dE at ...

#### Solution Summary

This solution provides a detailed step by step explanation of the given physics problem.