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Dipole moment of a non-uniform surface charge distribution

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Please refer to the attachment for solution.

SOLUTION
Y dPsinθ dP
B dθ dl
dPcosθ

r = asinθ

θ
C A X

Thin ring

D

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 ...

Solution Summary

Problem has been solved using integral calculus.

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