Question

Show derivation steps from equation (22-16) to (22-17) please show steps. Thank you.

the quantity s varies as we go through the eleme, remain the same, so we move them outside the integral. We find (22-15) 2rR 22-16) If the charge on the ring is negative, instead of positive as we have assumed, the This is a fine answer, but we can also switch to the total charge by using A-q (charged ring). magnitude of the field at P is still given by Eq. 22-16. However, the electric vector then points toward the ring instead of away from it. Let us check Eq 22-16 for a point on the central axis that is so far away that z>R. For such a point, the expression + R in Eq. 22-16 can be approximated as 22, and Eq. 22-16 becomes (charged ring at large distance). (22-17) 4πε 。 This is a reasonable result because from a large distolice, the ring "looks like" a point charge. If we replace z with r in Eq. 22-17, we iideed do have the magni- tude of the electric field due to a point charge, as given by Eq. 22-3. Let us next check Eq. 22-16 for a point at the center of the ring-that is, for z 0. At that point, Eq. 22-16 tells us that E-0. This is a reasonable result because if we were to place a test charge at the center of the ring, there would be no net electrostatic force acting on it; the force due to any element ot ring would be canceled by the force due to the element on the opposite siode the ring. By Eq.22-1, if the force at the center of the ring were zero, the electr field there would also have to be zero.

Answer 1

Given is:-

eq-1

Now,

consider only  

Let z= 99m and R= 0.1m and solving above expression we get

Now let z=99m and R=0m then

Therefore we can say that when z is far greater then R   we can assume R=0 for simple calculations therefore the whole expression eq-1 will become

by eliminating z from both numerator and denominator we get