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QJEEM22Ph96c4: JEE Main 2022 Physics - Class 12 JEEM22Ph96c4 Solution: Two point charges Q each are placed at a distance d apart. A…

Two point charges Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is :

Multiple Choice Options

  • A x = d
  • B x=d2x = {d \over 2}
  • C x=d2x = {d \over {\sqrt 2 }}
  • D x=d22x = {d \over {2\sqrt 2 }}

Step-by-Step Verified Solution

Force experienced by the charge q

F=kQqx[(d2)2+x2]32F = {{kQqx} \over {{{\left[ {{{\left( {{d \over 2}} \right)}^2} + {x^2}} \right]}^{{3 \over 2}}}}}

For maximum Coulomb's force for x

dFdx=0{{dF} \over {dx}} = 0

On solving x=d22x = {d \over {2\sqrt 2 }}

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