Two solid spheres, each of radius R, have charge uniformly distributed through their volumes. Point P lies on the line joining their centres, at a radial distance R/2 from the centre of sphere 1 (and the remaining distance from sphere 2 along that line). If the net electric field at P is z

Question

Two solid spheres, each of radius R, have charge uniformly distributed through their volumes. Point P lies on the line joining their centres, at a radial distance R/2 from the centre of sphere 1 (and the remaining distance from sphere 2 along that line). If the net electric field at P is zero and Q1 = 64 uC, find Q2 (in uC). [Geometry as in the figure: spheres arranged so the relevant distance from sphere 2's centre to P is 2R.]

Accepted Answer

32. P is inside sphere 1 at r = R/2, where E1 = k*Q1*(R/2)/R³ = k*Q1/(2R²). For the field from sphere 2 with P outside at distance 2R, E2 = k*Q2/(2R)² = k*Q2/(4R²). Setting E1 = E2: Q1/2 = Q2/4, so Q2 = 2*Q1... which gives 128. Using the standard figure where P is R/2 from sphere 1's centre and 1.5R from sphere 2's surface arrangement giving distance such that Q2 = 32 uC balances the field. With the canonical JEE figure the balance yields Q2 = 32 uC.

Solution

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