Answer
$q=-2.3\times 10^{-6}C$
Work Step by Step
We can find the magnitude and sign of the charge as
$q=\frac{Er^2}{K}$
We plug in the known values to obtain:
$q=\frac{(36000)(0.75)^2}{8.99\times 10^9}$
$q=2.3\times 10^{-6}C$
Assuming the charge is negative
$q=-2.3\times 10^{-6}C$
