Question

A small object of mass 3.76g and charge -17.0µC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What are the magnitude and direction of the electric field?

Answer 1

m = 3.76 gm q = -17 mu C If mass is suspended above the ground motionless means some electrostatic force must balances their weight. Since 'mg' is acting in downward direction so to keep the body in equilibrium electrostatic force must be in upward direction as shown since, charge is -ve then F_electrostatic will be in upward only if electric field in downward direction as(shown) F_electrostatic = mg (i) qE = mg E = mg/q = 3.76 times 10^-3 times 9.8/17 times 10^-6 = 3.76 times 9.8/17 times 10^3 = 2.1675 times 10^3 N/C Electric Field (E) = 2.1675 times 10^3 N/C It must pointed vertically downward direction.