Discussion Forum

Two charges, each equal to q are kept at x=-a  and x=a  on the x-axis. A particle of mass m and charge   q₀= q/2 is placed at the origin. If charges q₀  is given a small displacement y(y << a) along the y-axis,  the net force acting on the particle  is proportional to 

Answer:

Explanation:

$F_{net}=2F \cos\theta$

$F_{net}=\frac{2kq(\frac{q}{2})}{(\sqrt{y^{2}+a^{2})^{2}}}.\frac{y}{(\sqrt{y^{2}+a^{2})}}$

$F_{net}=\frac{2kq(\frac{q}{2})y}{(y^{2}+a^{2})^{3/2}}.\Rightarrow\frac{kq^{2}y}{a^{3}}\propto y$