Answer
Please see the work below.
Work Step by Step
(a) We know that
$ F_{net}=\frac{Kq_2}{2d^2}\sqrt{(q_1+q_3)^2+3(q_1-q_3)^2}$
This simplifies to:
$ d=\sqrt{\frac{Kq_2}{2F_{net}}\sqrt{(q_1+q_3)^2+3(q_1-q_3)^2}}$
We plug in the known values to obtain:
$ d=\sqrt{\frac{8.99\times 10^9N.m^2/C^2(6.3\mu C)}{2(0.65N)}\sqrt{(2.1\mu C+0.89\mu C)^2+3(2.1\mu C-0.89\mu C)^2}}$
$ d=40cm $
(b) We can find the required direction as follows:
$\theta=tan^{-1}[\frac{\sqrt 3(q_1-q_3)}{q_1+q_3}]$
We plug in the known values to obtain:
$\theta=tan^{-1}[\frac{\sqrt 3(2.1\times 10^{-6}-0.89\times 10^{-6})}{2.1\times 10^{-6}+0.89\times 10^{-6}}]$
$\theta=35^{\circ}$ in the counter-clockwise direction.
