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QJEEM20Ph8795: JEE Main 2020 Physics - Class 12 JEEM20Ph8795 Solution: A two point charges 4q and -q are fixed on the x-axis at x =…

A two point charges 4q and -q are fixed on the x-axis at x = −d2- {d \over 2} and x = d2{d \over 2} respectively. If a third point charge 'q' is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will : JEE Main 2020 (Online) 4th September Morning Slot Physics - Electrostatics Question 155 English

Multiple Choice Options

  • A increase by 3q24πε0d{{3{q^2}} \over {4\pi {\varepsilon _0}d}}
  • B increase by 2q23πε0d{{2{q^2}} \over {3\pi {\varepsilon _0}d}}
  • C decrease by q24πε0d{{{q^2}} \over {4\pi {\varepsilon _0}d}}
  • D decrease by 4q23πε0d{{4{q^2}} \over {3\pi {\varepsilon _0}d}}

Step-by-Step Verified Solution

ΔU=14πε0.4q.q(3d/2)−14πε0.4q.q(d/2)\Delta U = {1 \over {4\pi {\varepsilon _0}}}.{{4q.q} \over {(3d/2)}} - {1 \over {4\pi {\varepsilon _0}}}.{{4q.q} \over {(d/2)}}

=4q24πε0(2d)(−23)= {{4{q^2}} \over {4\pi {\varepsilon _0}}}\left( {{2 \over d}} \right)\left( { - {2 \over 3}} \right)

= decrease by 4q23πε0d{{4{q^2}} \over {3\pi {\varepsilon _0}d}}

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