Answer
$q=5.2\times10^{-20}C$.
Work Step by Step
The 2 dipole moments, $p_1$ and $p_2$. are equal in magnitude, and each one makes an angle of $52^{\circ}$ with the net dipole moment. We also know that $p_1=qd$, where q is the charge on the hydrogen atom and d is the distance between the hydrogen atom and the oxygen atom.
$$p=2p_1cos52^{\circ} $$
$$p_1=\frac{p}{2cos52^{\circ} }=qd$$
$$q=\frac{p}{2dcos52^{\circ} }=\frac{6.1\times10^{-30}C \cdot m}{2(0.96\times10^{-10}m)cos52^{\circ} }$$
$$q=5.2\times10^{-20}C$$
This about a third of an electron’s charge.
