Physics
posted by Anonymous on .
A charged particle of mass 6.38x10^31kg is suspended motionless in the air by an electric field. If the electric field is 7.52x10^21 N/C downwards, what is the magnitude and sign of the charge of
the particle?

Require than m * g = q * E
q is the charge. m is the mass in kg.
g is the acceleration of gravity, in m/s^2.
E is the field strength, in N/C
Solve for the field, charge, q, in Coulombs.
The charge must be negative so that the upward electrical force can balance the weight.