A water particle with a mass of 10.0 mg and a charge of 1.50 10⁻⁶ C remains suspended in a room. Determine the magnitude and direction of the electric field in the room.

Step-by-step solution

For the water particle to remain suspended, the upward electric force must balance the downward gravitational force. qE = mg E = mg q Substituting the given values, m = 10.0 mg = 10.0 10⁻⁶ kg , q = 1.50 10⁻⁶ C , and g = 9.8 m s ⁻² : E = 10.0 10⁻⁶ 9.8 1.50 10⁻⁶ E = 98 1.50 = 65.3 N C ⁻¹ Since the charge is positive, the electric field must be directed upward to exert an upward force.