JEE MAIN - Physics (2013 (Offline) - No. 18)
Two charges, each equals to $$q,$$ are kept at $$x=-a$$ and $$x=a$$ on the $$x$$-axis. A particle of mass $$m$$ and charge $${q_0} = {q \over 2}$$ is placed at the origin. If charge $${q_0}$$ is given a small displacement $$\left( {y < < a} \right)$$ along the $$y$$-axis, the net force acting on the particle is proportional to
$$y$$
$$-y$$
$${1 \over y}$$
$$-{1 \over y}$$
Explanation
_en_18_1.png)
$$ \Rightarrow {F_{net}} = 2F\,\cos \theta $$
$${F_{net}} = {{2kq\left( {{q \over 2}} \right)} \over {{{\left( {\sqrt {{y^2} + {a^2}} } \right)}^2}}}.{y \over {\sqrt {{y^2} + {a^2}} }}$$
$${F_{net}} = {{2kq\left( {{q \over 2}} \right)y} \over {{{\left( {{y^2} + {a^2}} \right)}^{3/2}}}} \Rightarrow {{k{q^2}y} \over {{a^3}}}$$
S0, $$F \propto y$$
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