Two charges, each equal to q , are kept at x = - a and x = a on the...

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

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

${F}_{net} = \dfrac{2kq\left(\dfrac{q}{2}\right)y}{{\left({y}^{2} + {a}^{2}\right)}^{3 / 2}} \approx \left(\dfrac{k{q}^{2}}{{a}^{3}}\right)y ⇒ F ∝ \left(- y\right)$

Net force $F$ is opposite the direction of displacement $y$. Hence, negative sign is used.