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QJEEM16Phae22: JEE Main 2016 Physics - Class 12 JEEM16Phae22 Solution: Within a spherical charge distribution of charge density (r), N …

Within a spherical charge distribution of charge density ρ\rho (r), N equipotential surfaces of potential V0, V0 + Δ\Delta V, V0 + 2 Δ\Delta V, .......... V0 + N Δ\Delta V ( Δ\Delta V > 0), are drawn and have increasing radii r0, r1, r2,..........rN, respectively. If the difference in the radii of the surfaces is constant for all values of V0 and Δ\Delta V then :

Multiple Choice Options

  • A ρ\rho (r) α\alpha r
  • B ρ\rho (r) = constant
  • C ρ\rho (r) α\alpha 1r{1 \over r}
  • D ρ\rho (r) α\alpha 1r2{1 \over {{r^2}}}

Step-by-Step Verified Solution

JEE Main 2016 (Online) 10th April Morning Slot Physics - Electrostatics Question 189 English Explanation

Here, Δ\Delta v and Δ\Delta r are same for any pair of surface.

we know,

Electric field, E = −- dvdr{{dv} \over {dr}}

∴\therefore    E = constant [As dv and dr are constant]

Electric field inside the spherical charge distribution.

E = ρr3ε0{{\rho r} \over {3{\varepsilon _0}}}

Now,   as E = constant

∴\therefore     ρ\rho  r = constant

⇒\Rightarrow     ρ\rho  (r) ∝\propto 1r{1 \over r}

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