Verified Solution • 2025–26 Session moderate

QJEEM20Pha3e4: JEE Main 2020 Physics - Class 12 JEEM20Pha3e4 Solution: A particle of charge q and mass m is subjected to an electric…

A particle of charge q and mass m is subjected to an electric field
E = E0 (1 – aa x2) in the x-direction, where aa and E0 are constants. Initially the particle was at rest at x = 0. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is :

Multiple Choice Options

  • A aa
  • B 2a\sqrt {{2 \over a}}
  • C 3a\sqrt {{3 \over a}}
  • D 1a\sqrt {{1 \over a}}

Step-by-Step Verified Solution

W=ΔKEW = \Delta KE

As inital and final kinetic energy both are zero so ΔKE\Delta KE = 0

∴\therefore W = 0

⇒\Rightarrow ∫0xFdx=0\int\limits_0^x {Fdx} = 0

⇒\Rightarrow q∫0xE0(1−ax2)dx=0q\int\limits_0^x {{E_0}\left( {1 - a{x^2}} \right)dx} = 0

⇒\Rightarrow qE0[∫0xdx−a∫0xx2dx]=0q{E_0}\left[ {\int\limits_0^x {dx - a} \int\limits_0^x {{x^2}dx} } \right] = 0

⇒\Rightarrow qE0[x−ax33]=0q{E_0}\left[ {x - {{a{x^3}} \over 3}} \right] = 0

⇒\Rightarrow x(1−ax23)=0x\left( {1 - {{a{x^2}} \over 3}} \right) = 0

⇒\Rightarrow x=0,x = 0, 1−ax23{1 - {{a{x^2}} \over 3}} = 0

⇒\Rightarrow ax23{{{a{x^2}} \over 3}} = 1

⇒\Rightarrow x=3a{x = \sqrt {{3 \over a}} }

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