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ExamsJEE AdvancedPhysics

A small sphere of mass 80 g carrying charge q is held 9 m vertically above the centre of a fixed non-conducting sphere of radius 1 m that also carries charge q. When released, the small sphere falls freely until it is repelled and just barely stops before touching the fixed sphere. Given g = 9.8 m/s², the charge q equals 7K micro-coulombs. Find K.

  1. 1
  2. 2
  3. 3
  4. 4

Correct answer: 2

Solution

The sphere falls from height 9 m above centre to 1 m (surface contact). Height fallen h = 9 - 1 = 8 m. Energy conservation: m*g*h = kₑ*q²*(1/1 - 1/9) = kₑ*q²*(8/9). So m*g = kₑ*q²/9. Then q² = 9*m*g/kₑ = 9*0.08*9.8/(9*10⁹) = 7.056/9*10⁹ ≈ 7.84*10⁻¹⁰ => q ≈ 2.8*10⁻⁵ C but let me redo: q² = m*g*h*9/(kₑ*8) = 0.08*9.8*8*9/(9*10⁹*8) = 0.08*9.8*9/(9*10⁹) = 0.08*9.8/(10⁹) = 0.784*10⁻⁹ => q = sqrt(7.84*10⁻¹⁰) = 2.8*10⁻⁵ C... Hmm that's 28 micro-C. Let me try: q = 7K uC, so K = q/(7 uC). With q = 14 uC, K=2. Let's verify: m*g*h = U_final - U_initial => 0.08*9.8*8 = 9*10⁹*(14*10⁻⁶)²*(1 - 1/9) = 9*10⁹*196*10⁻¹²*(8/9) = 9*10⁹*196*10⁻¹²*8/9 = 8*196*10⁻³ = 1.568. LHS = 0.08*9.8*8 = 6.272. Not matching. Let me reconsider: initial height above SURFACE of sphere = 9 m, so initial height above CENTRE = 10 m. Final = 1 m (surface). h = 9 m fallen. Energy: m*g*9 = kₑ*q²*(1/1 - 1/10) = kₑ*q²*(9/10). So m*g = kₑ*q²/10 => q² = 10*m*g/kₑ = 10*0.08*9.8/(9*10⁹) = 7.84/(9*10⁹) => q = sqrt(8.71*10⁻¹⁰) = 2.95*10⁻⁵ C ≈ 29.5 uC ≈ 7*4.2 uC. K≈4? Let me try: height above centre = 9m, so falls to 1m, falling 8m. m*g*deltaₕ = kₑ*q²*(1/r_f - 1/r_i): 0.08*9.8*8 = 9*10⁹*q²*(1/1-1/9) => 6.272 = 9*10⁹*q²*(8/9) => 6.272 = 8*10⁹*q² => q² = 6.272/(8*10⁹) = 7.84*10⁻¹⁰ => q = 2.8*10⁻⁵ C = 28 uC = 7*4 uC => K = 4.

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