Calculating Electric Field Above a Disc of Charge

[blueyellow]

Homework Statement

The electric field, E a distance z above a circular loop of charge density lambda, radius r, in the x-y plane centred on the origin, is given by

E(z)=[lambda z r] i(subscript z)/[2 epsilon0((z^2 + r^2)^(3/2))]

a) using this, find the electric field, E, a distance z above a disc of radius R and surface charge density sigma (8 marks)
b) describe the limits of E for the disc at R tends to infinity and z>>0 (4 marks)

The Attempt at a Solution

a) E(z)=[sigma z R] i(subscript z)/[2 epsilon0((z^2 + R^2)^(3/2))]

I don't think that's going to gain me the full 8 marks and I have no idea what to do next. Please help

b) I can't even make sense of it. How can z>>R if R tends to infinity?

 

[JesseC]

a) A disk of charge could be considered to be made up a large number of concentric loops of charge. You'd need to sum (think - integrate) the effects of many infinitesimally thin loops to get the field of a disk. To me, looks like you've just replaced r -> R and charge per unit length with charge per unit area which is wrong.

b) Question says z>>0, not z>>R.

 

[blueyellow]

but the question does say z>>R, I just checked

 

[blueyellow]

I made a mistake the first time I typed the question

 

[blueyellow]

never mind, I just found out that z>>R and z tends to infinity are two separate ques

 

[blueyellow]

E=[pi sigma z/[2 epsilon0]] ((1/z) -1/(sqrt(R^2 +z^2)))

so as R tends to infinity

E=p[i sigma/[2 epsilon0]] z(1/z -0)
=pi sigma/[2 epsilon0]??

 

[JesseC]

Looks plausible :) not going to thoroughly check it... you should try and convince yourself whether it is right or not. Does it seem sensible to you?