Calculating the Electric field for a ring

[link223]

Homework Statement

(II) A flat ring (inner radius R outer radius 4R ) is
uniformly charged. In terms of the total charge Q, determine the electric field on the axis at points (a) 0.25R and (b) 0.75R from the center of the ring. [Hint: The ring can be
replaced with two oppositely charged superposed disks.]

Relevant Equations

Gauss' Law

What i don't understand is why we are able to replace the ring with 'two oppositely charged superposed disks'?

Just trying to understand..
So we have a uniform charge which means that this'll just be a simplification of the problem than, correct?

Thanks in advance.

 

[Orodruin]

By the superposition principle, the solution to the sum of two charge distributions is the sum of the solutions to the separate charge distributions. If you take a small disk and a large disk with the opposite surface charge density (not charge!) then the sum of the charge distributions is an annulus as described in the problem as the resulting charge distribution within R is zero and between R and 4R there will only be the larger disc. Make sure to normalise your charge distribution such that the total charge distribution is of total charge Q.

 

[kuruman]

To me, the hint makes sense if the on-axis electric field due to a uniformly disk is assumed as given and there is no requirement to derive it. If there is such a requirement, I think doing the double integral would be simpler.

 

[Orodruin]

I think doing the double integral would be simpler.

Well, the double integral is also relatively easily written as a single integral just by symmetry arguments, making that approach even simpler. However, the difference in complexity is inserting zero twice in the primitive function (which evaluates to zero if you use the most convenient choice of integration constant).