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Electric field of a charge inside hollow conducting sphere

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(See attached file for full problem description)

A hollow grounded conducting sphere of radius a contains a point charge Q at the radius b as shown in the
figure (see attachment)

a) Show that the field inside the sphere is the same as if there were no sphere and, instead, a charge
Q' = -(a/b)Q at D = (a/b)a. You can prove this by showing that the V of Q plus Q' is uniform over the surface
of the sphere.
b) Calculate the force of attraction.
c) Calculate the surface charge density on the inside surface of the conducting sphere.

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The expert examines electric field of a charge inside hollow conducting sphere. The force of attraction are calculated.

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SOLUTION :

S

P

θ r1 r2
O Q R Q'

a b
a2/b

a) Let us consider an arbitrary point P on the surface of the sphere. Distance of point P from charge Q be r1 and that from Q' be r2. Let the radius vector OP make an angle θ with the line OQ'. Now, from the figure :

PR = OP sinθ = a sinθ and QR = OR - OQ = a cosθ - b

PQ2 = (r1)2 = PR2 + QR2 = (a sinθ)2 + (a cosθ - b)2 = a2sin2θ + a2cos2θ + b2 - 2abcosθ
__________________________ ______________
Or r1 = √a2(sin2θ + cos2θ) + b2 - 2abcosθ = √a2 + b2 - 2abcosθ [because sin2θ + cos2θ =1]
........(1)

Also, SQ' = OQ' sinθ = a2/b sinθ and PS = OS - OP = OQ' cosθ - OP = a2/b cosθ - a

PQ'2 = (r2)2 = PS2 + SQ'2 = (a2/b cosθ - a)2 + (a2/b sinθ)2 = [a2/b cosθ]2 + a2 - 2a3/b cosθ + [a2/b sinθ]2 = a4/b2 + a2 - 2a3/b cosθ = a2/b2 (a2 + b2 - 2ab cosθ)
____________________ ______________
Or r2 = √a2/b2 (a2 + b2 - 2ab cosθ) = a/b√a2 + b2 - 2ab cosθ ........(2)

Potential at point P due to two charges Q and Q' = V = Q/r1 + Q'/r2

Substituting Q' = -a/b Q and r1 and r2 from (1) and (2) we get :
...

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