Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by ρ(r) = kr, where r is the distance from the centre. Two charges A and B, of -Q each, are placed on diametrically opposite points, at equal distance, a, from the centre. If A and B do not experience any force, then :
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Question

Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by ρ(r) = kr, where r is the distance from the centre. Two charges A and B, of -Q each, are placed on diametrically opposite points, at equal distance, a, from the centre. If A and B do not experience any force, then :

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G1 Physics · Accepted Answer

The total charge inside the sphere is 2Q = 4π∫₀ᴿ kr·r²dr = πkR⁴, so k = 2Q/(πR⁴). For a < R, the enclosed charge within radius a is Qₑ = 4π∫₀ᵃ kr·r²dr = 2Qa⁴/R⁴. Thus the force on charge A due to the sphere has magnitude Fₛ = (1/4πε₀)(2Q²a²/R⁴). The repulsive force due to charge B, which is 2a away, is Fᵦ = (1/4πε₀)(Q²/4a²). For equilibrium, Fₛ = Fᵦ, giving 8a⁴ = R⁴. Therefore a = 8⁻¹ᐟ⁴R. Correct answer: C.