Final answer:
In the Millikan oil drop experiment, to find the magnitude of the electric force acting on a motionless drop, we use the net force equation Fnet = mdropg = qE. The electric force equals the drop's weight, 4×10^-14 newtons, because the drop is motionless.
Step-by-step explanation:
The scenario described is a classic illustration of the Millikan oil drop experiment, where the gravitational force on an oil drop is balanced by the electrical force when the drop is suspended motionless. To find the magnitude of the electric force acting on the drop, we can use the equation Fnet = mdropg = qE, where mdrop is the mass of the oil drop, g is the acceleration due to gravity (9.8 m/s2), q is the net charge on the oil drop, and E is the electric field.
Since the oil drop is suspended motionless, the magnitude of the electric force must exactly counteract the weight of the drop. Given that the weight of the drop (W) is 4×10-14 newtons (which is the force due to gravity), the electric force must also be 4×10-14 newtons. No additional information is provided to solve for the charge (q) or the electric field (E), but we know that their product must be equal to the weight of the drop.