Answer
a) $V_b-V_a=1.6\times10^4V$
b) $E=9.9\times10^4N/C$
$\theta=64^o$ above positive x axis
Work Step by Step
a) $V=k\frac{Q}{r}$
$V_b-V_a=k(3.8\mu C)\Big(\frac{1}{0.62m}-\frac{1}{0.88m}\Big)=1.6\times10^4V$
b) $E=k\frac{Q}{r^2}$
$E_a=k\frac{3.4\mu C}{(0.62m)^2}=8.90\times10^4N/C$
$E_b=k\frac{3.4\mu C}{(0.88m)^2}=4.42\times10^4N/C$
$E=\sqrt{E_a^2+E_b^2}=9.9\times10^4N/C$
$\theta=\arctan(\frac{8.90\times10^4N/C}{4.42\times10^4N/C})=64^o$ above positive x axis