A total charge Q is distributed uniformly over a rod of length L. The rod is aligned on the x- axis, with one end at the origin and the other at the point x = L. Calculate the field at point (L/2, D)

Answer 1

Due to symmetry, the x-component will cancel out.

Consider the left half of the rod.

Ey = E sin u = Int (k dq / r^2 sinu );

Here Int means integral.

dq = Q/L dx since it's uniformly distributed; sinu=D/r;

Plug into the Ey equation. Pull out the constant terms.

Ey=KQD/L Int(dx/r^3)

Here r = sqrt (D^2 + (L/2-x)^2)

Do the integration from 0 to L/2. Then double it.

Final answer Ey=2kQ/(D sqrt(4D^2+L^2))