Particle & Electromagnetism

Millikan's Oil-Drop Experiment

A charged oil drop hangs between two horizontal plates. Balance the weight force against the electric field, watch it drift to terminal velocity under air drag, and read the live force diagram.

Adjust the experiment

Charge on drop × e

Multiples of the elementary charge e = 1.60×10⁻¹⁹ C. Negative = extra electrons.

Drop mass ×10⁻¹⁵ kg

Tiny oil droplets, femtogram scale.

Plate voltage V

Positive drives the top plate positive (field points down).

Plate separation mm

Distance between the plates, d.

Air drag: ON Field lines: ON

Practice question

Press “New question” to begin.

Simulation

Field E = V/d

Terminal v

Weight force mg

Electric qE

Net force

Status

The charged drop moves under the combined forces; field lines run live between the plates.

Live force diagram

Weight (mg) Electric (qE)

Arrow lengths are proportional to force magnitude. When the two are equal and opposite, the drop is balanced.

The physics behind it

Three forces act on the drop. The weight force pulls down with Fg = mg. The electric force is Fe = qE = qV/d; its direction depends on the sign of the charge and the field. Air drag (Stokes' law) is Fd = 6πηrv and always opposes motion, which is why the drop settles to a steady terminal velocity instead of accelerating forever. When the electric force exactly balances the weight force, the drop floats still — the balance condition Millikan used to measure q = mgd / V. Use Auto-balance V to find the voltage that suspends the current drop. Drop radius is estimated from the mass assuming the density of oil (≈ 920 kg/m³), so drag responds realistically to size.