Answer
(a) $-1.14\times 10^5Nm^2/C$
(b) $1.46\times 10^5Nm^2/C$
(c) $3.28\times 10^5Nm^2/C$
(d) $-2.9\mu C$
Work Step by Step
(a) We know that the net charge is given as
$q=q_1+q_2$
$q=1.61\times 10^{-6}C-2.62\times 10^{-6}C=-1.01\times 10^{-6}C$
Now the electric flux can be determined as
$\phi=\frac{q}{\epsilon_{\circ}}$
$\phi=\frac{-1.01\times 10^{-6}C}{8.85\times 10^{-12}C^2/Nm^2}$
$\phi=-1.14\times 10^5Nm^2/C$
(b) The net charge is $q=q_2+q_3$
$\implies q=-2.62\times 10^{-6}C+3.91\times 10^{-6}C=+1.29\times 10^{-6}C$
Now $\phi=\frac{q}{\epsilon_{\circ}}$
$\phi=\frac{+1.29\times 10^{-6}C}{8.85\times 10^{-12}C^2/Nm^2}$
$\phi=1.46\times 10^5Nm^2/C$
(c) The net charge is $q=q_1+q_2$
$q=1.61\times 10^{-6}C-2.62\times 10^{-6}C+3.91\times 10^{-6}C$
$q=2.90\times 10^{-6}C$
Now $\epsilon=\frac{q}{\epsilon_{\circ}}$
$\epsilon=\frac{2.90\times 10^{-6}C}{8.85\times 10^{-12}C^2/Nm^2}$
$\epsilon=3.28\times 10^5Nm^2/C$
(d) We know that if want to have zero flux, then we have to add the fourth charge, which is equal to the total charge of $q_1$, $q_2$ and $q_3$ but opposite in sign.
That is, $Q=-(q_1+q_2+q_3)$
$\implies Q=-2.9\mu C$
