Question

Oil Drop 11. Robert Millikan received a Nobel Prize for determining the charge on the electron. To do this, he set up a of oil between potential difference between two horizontal parallel metal plates. He then sprayed drops the plates and adjusted the potential difference until drops of a certain size between the plates, as shown above. Suppose that when the potential difference adjusted until the electric field is 10,000 N/C downward, a certain drop with a mass of remains suspended remained suspended at rest between the plates is 3.27 x 10-1e kg a. What is the magnitude of the charge on this drop? b. The electric field is downward, but the electric force on the drop is upward. Explain why. If the distance between the plates is 0.01 m, what is the potential difference between the plates? c. The oil in the drop slowly evaporates while the drop is being observed, but the charge on the drop remains the same. Indicate whether the drop remains at rest, moves upward, or moves downward. Explain briefly. d.

Answer 1

Forces acting on drop are electrostatic force = q E upwards gravitation force = mg downwards q E = mg = > q = mg/E q = 3.27 times 10^-16 times 9.8/10.000 = 32.046 times 10^-20 c electrostatic force acting on drop in vector from F^vector = q E^vector here E^vector is downward F^vector is upward that means charge on drop must be negative. Therefore force on negative charge opposite electric field. E = v/d = > V = Ed = 10000 times 0.01 V = 100v electrostatic force on drop remains same because q is same. But when oil in the drop evaporates its mass decrees so mg decreases then drop more upwards.