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JEE Main 2014 (12 Apr, Shift 1) · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from JEE Main 2014 (12 Apr, Shift 1)

Q116JEE Main 2014 · 12 APR · SHIFT 1Difficult

Electric Flux & Gauss's Law

A spherically symmetric charge distribution is characterised by a charge density having the following variation:

ρ\rho(r) = ρ\rho0 (1−rR)\left(1 - \frac{r}{R}\right) for r < R

ρ\rho(r) = 0 for r ≥\geq R

Where r is the distance from the center of the charge distribution and ρ\rho0 is a constant. The electric field at an internal point (r < R) is:

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Answer: (b)

  1. Enclosed charge: q=4πρ0∫0r(r2−r3R)dr=4πρ0(r33−r44R)q = 4\pi\rho_0\int_0^r\left(r^2 - \frac{r^3}{R}\right)dr = 4\pi\rho_0\left(\frac{r^3}{3} - \frac{r^4}{4R}\right)
  2. E=q4πε0r2=ρ0ε0(r3−r24R)E = \frac{q}{4\pi\varepsilon_0 r^2} = \frac{\rho_0}{\varepsilon_0}\left(\frac{r}{3} - \frac{r^2}{4R}\right)
  3. So the answer is (b).
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