Answer
$k = 4500~N/m$
Work Step by Step
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:
$ky = F$
$k = \frac{F}{y}$
$k = \frac{225.4~N}{0.050~m}$
$k = 4500~N/m$
