Question

6. a) Applying the superposition' principle,find an expression vector at the distance X from the center of the ring. for the electric field b) Calculate the electric charge on a ring with radius 18 cm, f distance of 20 cm from its center is 25 KN/C

Answer 1

By Pythagoras theorem, OP = squareroot R^2 + x^2 let the ring has linear charge density 'lambda' field due to 'dl' at P therefore dE = 1/4 pi epsilon_0 lambda dl/(R^2 + x^2) {because lambda = dq/dl} Now, the component of dE^bar, which is acting downwards (dE sin theta) will be cancelled by the second length 'dl_2' which is vertically opposite is 'dl' so, E^vector = integral dE = integral 1/4 pi epsilon_0 lambda dl/(R^2 + x^2) cos theta {because cos theta = x/(R^2 + x^2)^1/2} E^vector = integral 1/4 pi epsilon_0 lambda dl x/(R^2 + x^2)^1/2 integrating over total length {integral dl = 2 pi R} E^vector = 1/4 pi epsilon_0 x/(R^2 + x^2)^3/2 Radially outwards R = 0.18 m, x = 0.2 m, E = 25 times 10^3 N/c: 1/4 pi epsilon_0 k = 9 times 10^9 putting the values is above equations 25 times 10^3 = 9 times 10^9 times Q times 0.2/(0.18^2 + 0.2^2)^3/2