Answer
The spring constant is $~~8900~N/m$
Work Step by Step
We can write Coulomb's Law:
$F = \frac{k~q_1~q_2}{r^2}$
We can find the spring constant $k'$:
$k'x = \frac{k~q_1~q_2}{r^2}$
$k' = \frac{k~q_1~q_2}{x~r^2}$
$k' = \frac{(9.0\times 10^9~N~m^2/C^2)~(2.0\times 10^{-6}~C)(4.0\times 10^{-6}~C)}{(0.012~m)~(0.026~m)^2}$
$k' = 8900~N/m$
The spring constant is $~~8900~N/m$
