Answer
(a) $76.7KV $
(b) $14.1m/s $
Work Step by Step
(a) We can find the distances $ r_1, r_3$ as
$ r_1,r_3=\frac{1.25m}{2}=0.625m $
and $ r_2=\sqrt{(1.25m)^2-(0.625m)^2}=1.083m $
Now $ V=K(\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3})$
We plug in the known values to obtain:
$ V=(8.99\times 10^9N.m^2/C^2)[\frac{2.75}{0.625m}+\frac{7.45}{1.083m}+\frac{(-1.72)}{0.625m}]\times 10^{-6}C=76.7KV $
(b) According to the law of conservation of energy
$ K_i+U_i=K_i+U_f $
$\implies 0+q_4V=\frac{1}{2}mv_f^2+0$
This simplifies to:
$ v_f=\sqrt{\frac{2q_4V}{m}}$
We plug in the known values to obtain:
$ v_f=\sqrt{\frac{2(6.11\times 10^{-6}C)(7.67\times 10^3V)}{4.71\times 10^{-3}Kg}}$
$ v_f=14.1m/s $