Charged Sphere Between Deflecting Plates
A tiny oil droplet carrying a net charge of −3.2 x 10^-19 C is sprayed into the region between two horizontal parallel plates separated by 8.0 mm. The droplet enters with a horizontal speed of 2.5 m/s and is observed to travel in a straight, horizontal line when the potential difference across the plates is 1.2 kV. The student then doubles the potential difference and observes the droplet curve. The droplet mass is 2.4 x 10^-15 kg. Gravitational field strength is 9.8 N/kg, and the elementary charge is 1.60 x 10^-19 C. The student claims the straight-line motion proves there is no electric force on the droplet.
1. Evaluate the student's claim and determine which plate is positive.
Read the worked answer
The straight-line horizontal motion means the net vertical force is zero, not that the electric force is zero. Gravity acts downward with magnitude mg = (2.4e-15)(9.8) = 2.35e-14 N. For the droplet to move horizontally, the electric force must be upward with the same magnitude.
The droplet is negative, so the electric force points opposite the field. An upward force on a negative charge requires the field to point downward, which means the top plate is positive and the bottom plate is negative.
Check the magnitude: E = V/d = 1200/0.0080 = 1.5e5 V/m. Then qE = (3.2e-19)(1.5e5) = 4.8e-14 N. This is about twice mg, so the quoted voltage would actually produce an upward acceleration, not balanced motion. The student's claim is wrong, and the data are internally inconsistent: the balancing voltage should be V = mgd/q = (2.35e-14)(0.0080)/(3.2e-19) ≈ 590 V. A good answer flags this discrepancy rather than accepting the 1.2 kV figure.
2. Assuming the droplet is initially balanced at the voltage that truly balances it, describe quantitatively what happens after the voltage is doubled.
Read the worked answer
At balance, qE = mg. Doubling the voltage doubles E, so the upward electric force becomes 2mg while gravity remains mg. The net upward force is mg, giving acceleration a = mg/m = g = 9.8 m/s^2 upward.
The droplet crosses the 8.0 mm gap in time t from d = ½at^2: t = sqrt(2d/a) = sqrt(2 × 0.0080/9.8) ≈ 0.040 s. During that time its horizontal displacement is vx t = (2.5)(0.040) ≈ 0.10 m, so it strikes the upper plate about 10 cm downstream of where it entered (assuming the plates are that long).
The key judgement is that the vertical acceleration equals g, independent of the droplet's mass, because both forces scale with the same quantities. Students who plug in q, E, and m separately should reach the same 9.8 m/s^2.