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JEE Main 2010 (AIEEE) · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from JEE Main 2010 (AIEEE)

Q123JEE Main 2010 · AIEEEDifficult

Electric Flux & Gauss's Law

Let there be a spherically symmetric charge distribution with charge density varying as ρ(r)=ρ0(54−rR)\rho(r) = \rho_0\left(\frac{5}{4} - \frac{r}{R}\right) upto r=R,r = R, and ρ(r)=0\rho(r) = 0 for r>R,r > R, where rr is the distance from the origin. The electric field at a distance r (r<R)r\ (r < R) from the origin is given by

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Answer: (a)

  1. Enclosed charge: q=4πρ0∫0r(54r2−r3R)dr=4πρ0(5r312−r44R)=πρ0r3(53−rR)q = 4\pi\rho_0\int_0^r\left(\frac{5}{4}r^2 - \frac{r^3}{R}\right)dr = 4\pi\rho_0\left(\frac{5r^3}{12} - \frac{r^4}{4R}\right) = \pi\rho_0 r^3\left(\frac{5}{3} - \frac{r}{R}\right)
  2. E=q4πε0r2=ρ0r4ε0(53−rR)E = \frac{q}{4\pi\varepsilon_0 r^2} = \frac{\rho_0 r}{4\varepsilon_0}\left(\frac{5}{3} - \frac{r}{R}\right)
  3. So the answer is (a).
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