JEE MAIN - Physics (2022 - 29th June Evening Shift - No. 3)
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 :
x = d
$$x = {d \over 2}$$
$$x = {d \over {\sqrt 2 }}$$
$$x = {d \over {2\sqrt 2 }}$$
Explanation
Force experienced by the charge q
$$F = {{kQqx} \over {{{\left[ {{{\left( {{d \over 2}} \right)}^2} + {x^2}} \right]}^{{3 \over 2}}}}}$$
For maximum Coulomb's force for x
$${{dF} \over {dx}} = 0$$
On solving $$x = {d \over {2\sqrt 2 }}$$
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