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)?

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:

from which we find q:

Wouldn't it be 0, because if the frog is at 180 and it moves 18 back, it would be 0, 6 + 12 = 18 So the Answer is 0