The disc can be regarded as to be composed of several infinitesimally thin rings. Let us consider one such ring of radius r and thickness dr. The charge contained in this ring will be,
Using the expression derived for the electric field due to a ring in Problem 3.9 Eqn 1, the electric field due to the infinitesimally thin ring is give by,
Similarly using the expression derived in Problem 3.30 Eqn 1, we determine the potential as,
When l tends to 0, From (2) and (3),
When l much greater than R, using Taylor's series,
Using (4a) and (4b) respectively on (2) and (3) we obtain,
why aren't we taking the vertical component for the field.. as cos theta = l/sqrt(l^2+r^2)
ReplyDeleteWe do, not explicitly though. It says "Using the expression derived for the electric field due to a ring in Problem 3.9 Eqn 1, the electric field due to the infinitesimally thin ring is give by,"
DeleteYou can check that problem for an explanation of where does the expression come.
Eventhough, it's true that in the differential expressions for the Electric field they forgot to introduce an L, but in the final expression they introduced it to get the right answer