Answer
$a) E = 720N/C $ downward towards the negative charge.
$b) r = 1.936m$
Work Step by Step
$q = -5nC$
a) $E = {kq\over r^2} = {9\times 10^9 \times 5\times 10^{-9}\over (0.250)^2} = 720N/C$
and directed toward negative charge i.e downwards
$b) r = \sqrt{\frac{kq}{E}} = \sqrt{\frac{9\times 10^9 \times 5\times 10^{-9}}{12}} = 1.936m$
