See attached file for full problem description.
Please refer to the attachment for solution.
Y dPsinθ dP
B dθ dl
r = asinθ
C A X
Charge density on the sphere = σ = σ0 cosθ ...........(1)
As the charge density varies as per (1) above, its value will be + σ0 at point A and - σ0 at point C with nil value at the circumference of the circle passing through points B and D. Thus the right hemisphere will be positively charged and the left one negatively charged.
Let us now consider a point on the surface of the sphere whose position vector is a θ (a is the radius of the sphere and θ is the angle made by the position vector with X axis). If we take the locus of all such points on the surface (all points which make an angle θ with X axis) we get a circle on the surface with radius ...
Problem has been solved using integral calculus.