Question

What must the charge (sign and magnitude) of a particle of mass 1.42 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 610 n/c ? use 9.81 m/s2 for the magnitude of the acceleration due to gravity. view available hint(s)?

Answer 1

We have two forces acting on the particle: the weight of the particle, downward, with intensity W=mg, and the Coulomb's force due to the electric field, with intensity F=qE. 

In order to keep the particle in equilibrium, F must point upward. The direction of F depends on the sign of the charge. The electric field's direction is downward, so if we want F to point upward, the charge q must have negative sign.

Then, to find the magnitude of the charge, we should require that the intensity of the two forces acting on the particle is equal:
$mg=qE$
from which we find q:
$q= \frac{mg}{E} = \frac{(1.42 \cdot 10^{-3}kg)(9.81 m/s^2)}{610 N/C}=2.28 \cdot 10^{-5}C$