Answer
$q_3=-7.3\mu C$
Work Step by Step
We are given that at the origin, the electric potential is zero
$\implies \frac{1}{4\pi \epsilon_{\circ}}(\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3})=0$
$\implies \frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}=0$
We plug in the known values to obtain:
$\frac{24.5\times 10^{-6}C}{7.61m}+\frac{-11.2\times 10^{-6}C}{8.13m}+\frac{q_3}{3.99m}=0$
This simplifies to:
$q_3=-7.3\mu C$
