Purchase Solution

# Electric field due to a charged disc

Not what you're looking for?

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.

##### Solution Summary

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

##### 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 θ and θ+dθ with the X axis. Area of the element = rdθdr. Charge on the element = dq = σ rdθdr where σ 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 = (1/4ΠЄ0)dq/l2 = (1/4ΠЄ0)σ rdθdr/l2

From the fig. l2 = s2+r2. Hence, dE = (1/4ΠЄ0)σ rdθdr/(s2+r2) .......(1)

The electric field vector dE at ...

##### The Moon

Test your knowledge of moon phases and movement.

##### Intro to the Physics Waves

Some short-answer questions involving the basic vocabulary of string, sound, and water waves.

##### Variables in Science Experiments

How well do you understand variables? Test your knowledge of independent (manipulated), dependent (responding), and controlled variables with this 10 question quiz.