Question 180 of 194

Two charges, each equal to $\text{q}$, are kept at $\text{x}=-\text{a}$ and $\text{x}=\text{a}$ on the x-axis. A particle of mass $\text{m}$ and charge ${\text{q}}_{0}=-\dfrac{\text{q}}{2}$ is placed at the origin. If charge ${\text{q}}_{0}$ is given a small displacement $\text{(y \lt \lt a)}$ along the y-axis, the net force acting on the particle is proportional to :

A$\dfrac{1}{\text{y}}$B$-\dfrac{1}{\text{y}}$C$\text{-y}$D$\text{y}$

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Answer: (C) $\text{-y}$

$\Rightarrow {\text{F}}_{\text{net}}=\text{2Fcos}\theta$

${F}_{net}=\dfrac{2kq\left(\dfrac{q}{2}\right)}{{\left(\sqrt{{y}^{2}+{a}^{2}}\right)}^{2}}\left(\dfrac{y}{\sqrt{{y}^{2}+{a}^{2}}}\right)$

${F}_{net}=\dfrac{2kq\left(\dfrac{q}{2}\right)y}{{\left({y}^{2}+{a}^{2}\right)}^{3/2}}\approx \left(\dfrac{k{q}^{2}}{{a}^{3}}\right)y \Rightarrow F\propto \left(-y\right)$
Net force $F$ is opposite the direction of displacement $y$. Hence, negative sign is used.