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JEE Main20199 Jan 2019Evening ShiftPhysicsElectrostaticsActual

Charge is distributed within a sphere of radius R with a volume charge density ρ r = A r 2 e - 2 r a , where A and a are constants. If Q is the total charge of this charge distribution, the radius R is:

Options

  1. Aa 2 log ⁡ 1 1 - Q 2 π a A
  2. Ba log ⁡ 1 1 - Q 2 π a A
  3. Ca log ⁡ 1 - Q 2 π a A
  4. Da 2 log ⁡ 1 - Q 2 π a A

Correct answer

A. a 2 log ⁡ 1 1 - Q 2 π a A

Step-by-step solution

ρ r = A r 2   e - 2 r a Charge enclosed between r and r + d r is d q = ρ r   4 π r 2 d r To get total charge Q Q = ∫ 0 R d q = ∫ 0 R A r 2 e - 2 r / a   4 π r 2 d r = - 4 πA a 2 [ e - 2 r / a ] 0 R Q = - 4 π   A a 2 e - 2 R / a - 1 R = - ln ⁡ 1 - Q 2 4 πAa a 2 = a 2 ln ⁡ 1 1 - Q 2 πaA

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