Electromagnetism

Question

A ring of radius R has a homogeneous charge distribution. Place the origin at the center of the ring and calculate the electrostatic field for any point r such that |r| < R.

Answer 1

Due to given distribution charge dq of an elemenr dl of the ring, equals to dq=Qdl/2πR;

then dE(r )=dq/(4πεr^2), where    r^2=R^2 +x^2 and cosθ=x/(R^2+x^2)^1/2.

At any point P along х-axis (center of the ring as origin) you have Y and X components of the vector E. 

Due to symmetry sum of Y components gives zero.

Thus, you need to consider only x component dEx=cosθdE.

Then after integration for E(r )=Qx/[4πε(R^2 +x^2)^3/2]