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JEE Main 2019 (9 Jan, Shift 2) · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from JEE Main 2019 (9 Jan, Shift 2)

Q106JEE Main 2019 · 9 JAN · SHIFT 2Difficult

Electric Flux & Gauss's Law

Charge is distributed within a sphere of radius R with a volume charge density ρ(r)=Ar2e−2ra,\rho(r) = \frac{A}{r^{2}}e^{-\frac{2r}{a}}, where A and a are constants. If Q is the total charge of this charge distribution, the radius R is:

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Answer: (b)

  1. Q=∫0RAr2e−2r/a4πr2dr=4πA⋅a2(1−e−2R/a)=2πaA(1−e−2R/a)Q = \int_0^R \frac{A}{r^2}e^{-2r/a} 4\pi r^2 dr = 4\pi A \cdot \frac{a}{2}\left(1 - e^{-2R/a}\right) = 2\pi aA\left(1 - e^{-2R/a}\right)
  2. e−2R/a=1−Q2πaAe^{-2R/a} = 1 - \frac{Q}{2\pi aA}
  3. R=a2log⁡(11−Q2πaA)R = \frac{a}{2}\log\left(\frac{1}{1 - \frac{Q}{2\pi aA}}\right)
  4. So the answer is (b).
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