Answer
$Q=1.37\times 10^{-10}C$
Work Step by Step
The net electric field is given as
$E_{net}=E_{Q^+}+E_{Q^{-}}=2E$.........eq(1)
But we know that:
$E=K\frac{Q}{r^2}$ in the given scenario $r=\frac{r}{2}$
So,
$E=K\frac{Q}{(\frac{r}{2})^2}$
Thus the equation(1) becomes:
$E_{net}=2(K\frac{Q}{(\frac{r}{2})^2})$
This simplifies to
$E_{net}=\frac{8KQ}{r^2}$
This can be rearranged as
$Q=\frac{E_{net}r^2}{8K}$
We plug in the known values to obtain:
$Q=\frac{386(0.16)^2}{8(8.99\times 10^9)}=1.37\times 10^{-10}C$
