Verified Solution • 2025–26 Session moderate

QJEEM18Phcbdc: JEE Main 2018 Physics - Class 12 JEEM18Phcbdc Solution: A solid ball of radius R has a charge density given by = o (1 r …

A solid ball of radius R has a charge density ρ\rho
given by ρ\rho = ρ\rho o (1 −- r/R) for 0 ≤\le r ≤\le R. The electric field outside the ball is :

Multiple Choice Options

  • A ρoR3∈or2{{{\rho _o}{R^3}} \over {{ \in _o}{r^2}}}
  • B ρoR312∈or2{{{\rho _o}{R^3}} \over {12{ \in _o}{r^2}}}
  • C 4ρoR33∈or2{{4{\rho _o}{R^3}} \over {3{ \in _o}{r^2}}}
  • D 3ρoR34∈or2{{3{\rho _o}{R^3}} \over {4{ \in _o}{r^2}}}

Step-by-Step Verified Solution

Electric field outside the ball is given by

E = 14πε0qr2{1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}} .............(i)

Now, dq = ρ\rho dV = ρ\rho (4 π\pi r2)dr

∴\therefore q = ∫dq=∫0Rρ0(1−rR)(4πr2)dr\int {dq = \int\limits_0^R {{\rho _0}\left( {1 - {r \over R}} \right)\left( {4\pi {r^2}} \right)dr} }

= (4πρ0)[r33−1R×r44]0R{\left( {4\pi {\rho _0}} \right)\left[ {{{{r^3}} \over 3} - {1 \over R} \times {{{r^4}} \over 4}} \right]_0^R}

= (4πρ0)(R33−R34){\left( {4\pi {\rho _0}} \right)\left( {{{{R^3}} \over 3} - {{{R^3}} \over 4}} \right)}

⇒\Rightarrow q = (4πρ0)(R312){\left( {4\pi {\rho _0}} \right)\left( {{{{R^3}} \over {12}}} \right)} ......(ii)

From eqns. (i) and (ii), E = ρoR312∈or2{{{\rho _o}{R^3}} \over {12{ \in _o}{r^2}}}

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