Verified Solution • 2025–26 Session moderate

QJEEM19Ph637c: JEE Main 2019 Physics - Class 12 JEEM19Ph637c Solution: Four point charges –q, +q, +q and –q are placed on y-axis at y…

Four point charges –q, +q, +q and –q are placed on y-axis at y = –2d, y = –d, y = +d and y = +2d, respectively. The magnitude of the electric field E at a point on the x-axis at x = D, with D >> d, will behave as :-

Multiple Choice Options

  • A E∝1D3E \propto {1 \over D^3}
  • B E∝1DE \propto {1 \over D}
  • C E∝1D4E \propto {1 \over D^4}
  • D E∝1D2E \propto {1 \over D^2}

Step-by-Step Verified Solution

Electric field at p = 2E1cos θ\theta 1 –2E2cos θ\theta 2

= 2Kq(d2+D2)×D(d2+D2)1/2−2Kq[(2d)2+D2]×D[(2d)2+D2]1/2{{2Kq} \over {\left( {{d^2} + {D^2}} \right)}} \times {D \over {{{\left( {{d^2} + {D^2}} \right)}^{1/2}}}} - {{2Kq} \over {\left[ {{{\left( {2d} \right)}^2} + {D^2}} \right]}} \times {D \over {{{\left[ {{{\left( {2d} \right)}^2} + {D^2}} \right]}^{1/2}}}}

=2KqD[(d2+D2)−3/2−(4d2+D2)−3/2]= 2KqD\left[ {{{\left( {{d^2} + {D^2}} \right)}^{ - 3/2}} - {{\left( {4{d^2} + {D^2}} \right)}^{ - 3/2}}} \right]

=2KqDD3[(1+d2D2)−3/2−(1+4d2D2)−3/2]= {{2KqD} \over {{D^3}}}\left[ {{{\left( {1 + {{{d^2}} \over {{D^2}}}} \right)}^{ - 3/2}} - {{\left( {1 + {{4{d^2}} \over {{D^2}}}} \right)}^{ - 3/2}}} \right]

Applying binomial approximation ∵\because d << D

=2KqDD3[1−32d2D2−(1−3×4d22D2)]= {{2KqD} \over {{D^3}}}\left[ {1 - {3 \over 2}{{{d^2}} \over {{D^2}}} - \left( {1 - {{3 \times 4{d^2}} \over {2{D^2}}}} \right)} \right]

=2KqDD3[122d2D2−32d2D2]= {{2KqD} \over {{D^3}}}\left[ {{{12} \over 2}{{{d^2}} \over {{D^2}}} - {3 \over 2}{{{d^2}} \over {{D^2}}}} \right]

=9kqd2D4= {{9kq{d^2}} \over {{D^4}}}

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