Question

Consider an annular disc of inner radius r and outer radius 2r, having uniform surface charge density . The electric field on the axis of this disc at a distance r from the centre will be

Answer 1

Answer:

Such a surface charge density is conventionally given the symbol sigma. For a disk, we have the relationship

where Q is the total charge and R is the radius of the disk. A ring of thickness da centered on the disk as shown has differential area given by

and thus a charge given by

The field produced by this ring of charge is along the x-axis and is given by the previous result:

The total field is given by simply integrating over a from 0 to R

The integral is actually 'perfect' and is given by

After substituting the limits, we get the final result:

Very far from the disk, we need to use a series approximation with x much larger than R. The algebra is in Tipler, but rest assured that we simply recover the simple Coulomb law result

Answer 2

Explanation:

The disk has a uniform positive surface charge density δ ... I believe I have to use a dh (height) term in here .... You should have rings with inner radius R, and outer .