A water particle with a mass of 10.0 mg and a charge of 1.50 × 10^...

For the water particle to remain suspended, the upward electric force must balance the downward gravitational force.

$qE = mg$

$E = \dfrac{mg}{q}$

Substituting the given values, $m = 10.0 \text{ mg} = 10.0 \times 10^{-6} \text{ kg}$, $q = 1.50 \times 10^{-6} \text{ C}$, and $g = 9.8 \text{ m s}^{-2}$:

$E = \dfrac{10.0 \times 10^{-6} \times 9.8}{1.50 \times 10^{-6}}$

$E = \dfrac{98}{1.50} = 65.3 \text{ N C}^{-1}$

Since the charge is positive, the electric field must be directed upward to exert an upward force.