College Physics (4th Edition)

$k = 4500~N/m$
We can find the angle between the vertical and the line between the $7.0~\mu C$ charge and each $-4.0~\mu C$ charge: $tan~\theta = \frac{2.0~cm}{4.0~cm}$ $\theta = tan^{-1}(\frac{2.0~cm}{4.0~cm})$ $\theta = 26.565^{\circ}$ The horizontal component of the net electric force on the $7.0~\mu C$ charge is zero. We can find the magnitude of the net electric force exerted downward on the $7.0~\mu C$ charge due to the other two charges: $F_y = 2\times \frac{k~\vert q_1 \vert~\vert q_2 \vert}{r^2}~cos~\theta$ $F_y = 2\times \frac{(9.0\times 10^9~N~m^2/C^2)(7.0\times 10^{-6}~C)(4.0\times 10^{-6}~C)}{\left(\sqrt{(0.020~m)^2+(0.040~m)^2~~}~~\right)^2}~(cos~26.565^{\circ})$ $F_y = 2\times \frac{(9.0\times 10^9~N~m^2/C^2)(7.0\times 10^{-6}~C)(4.0\times 10^{-6}~C)}{0.002~m^2}~(cos~26.565^{\circ})$ $F_y = 225.4~N$ We can find the spring constant: $ky = F$ $k = \frac{F}{y}$ $k = \frac{225.4~N}{0.050~m}$ $k = 4500~N/m$