Question

Show steps of derivation from equation (22-26) to (22-27) please include explanations. Thank you.

where we have pulled the constants (including z) out of the integral. T this integral, wecast it in the form f X ndX by setting X = (z2 + r2). )o solve and dx (2r) dr. For the recast integral we have m+ 1 and so Eq. 22-24 becomes (22-25) 0 Taking the limits in Eq. 22-25 and rearranging, we find (22-26) 2e(charged disk) as the magnitude of the electric field produced by a flat, circular, charged disk a 20. points on its central axis. (In carrying out the integration, we assumed that z If we let Roo while keeping z finite, the second term in the parenthes Eq 22-26 approaches zero, and this equation reduces to es in E280 (22-27) (infinite sheet). This is the electric field produced by an infinite sheet of uniform charge located on one side of a nonconductor such as plastic. The electric field lines for such a situation are shown in Fig. 22-4. We also get Eq. 22-27 if we let z → 0 in Eq. 22-26 while keeping R finite·This shows that at points very close to the disk, the electric field set up by the disk is the same as if the disk were infinite in extent.

Answer 1

Given is:-

eq-1

Now,

If R tends to infinity then it means R is a very very very large number as compare to z therefore the whole denominator will become very large and will result in zero.

For example

let z=10 and R=999999999

we get

or in simple terms

by plugging this value in eq-1 then the second term in the parantheses will become zero

which gives us