Dalia,

This one is a bit of a stretch to explain without being able to diagram as I, but I'll describe the solution, and you draw the sketch, OK?

!st let's agree to have the plane of the disk coincide with the x-y plane with it's center at the origin. This will make the axis through the center of the disk, and perpendicular to it the z-axis. We wish to find an expression for the electric field at some point P along the z-axis at a distance "z" from the origin. Draw a disk with radius R in the x-y plane with the z-axis through the center and label your point P.

The electric field is defined by E_{r}** **= kq/r². We are not given the total charge q on the disk, but we are given the charge density σ and the total charge will be q = σ·A, where A = 2πr².

Now draw a second circle on the disk and about the origin, some dist r away from the origin such that 0<r<R.

Now draw one more circle slightly larger than this at a distance r+dr. This gives us a circular ring of charge a distance r away from the origin which has a length of 2πr and width dr. The area of this ring of charge is going to be: dA = 2πrdr.

We're going to need to integrate over the surface of the disk from r = 0 to r = R to get the total contribution form all the surface charge elements to get the field at P. We can do this by noting that:

dE = kdq/r², but dq = σ·dA = 2πrσdr

So we can now write: dE = 2πkσrdr/r²

But the r we've written here is for the radius of the disk from the origin out to where r = R. We need to relate this to the distance to our point P on the z-axis.

We first note that the distance z from the origin to point P is at 90° to the plane of the disk and therefore at 90° to our radius vector r which points to the small charge element dq which lies at the location of our ring element dr on the disk. The distance from this element dq to our point P is the length of the hypotenuse of the right triangle we just described: so our r² in the denominator becomes: (z²+r²).

We still have an r in the numerator to consider. Think about what happens as be keep drawing lines from the circumference of our circle to the point P. All of the components dE of our electric field cancel in the x-y plane, leaving only the components rcosθ which parallel the z-axis. (θ is the angle at point P between the z-axis and the line from charge element at dr to point P.

rcosΘ is then = z/(z²+r²)½. We can now write that

dE = 2πkσz·(z²+r²)^{-3/2} dr

I will leave it to you to perform the integration from r = 0 to R, but you should get:

E = 2πkσ[1-z/(z²+R²)½]

Plug in you values for R and z to complete your solutions.

Dalia S.

09/16/14