How does the balance method work?

First hold the drop still: qU/d = mg. Then switch the field off; drag quickly balances gravity and the drop falls at constant speed, so Stokes’ law mg = 6πηav_g gives the radius a = √(9ηv_g/2ρg) and the mass m = (4/3)πa³ρ. Combined, q = 18π/√(2ρg)·[ηl/(t_g(1 + b/pa))]^{3/2}·d/U, so only the balance voltage U and fall time t_g are needed.

How is the dynamic method different?

The drop need not be held still. A larger voltage makes it rise at constant speed and you time the rise: qU/d − mg = 6πηav_e. Together with mg = 6πηav_g for the fall this gives q = (mgd/U)(v_g + v_e)/v_g. It avoids the difficulty of setting the balance voltage exactly, at the cost of one more timing.

Why is the Cunningham correction needed?

A drop about 1 µm across is not much bigger than the mean free path of air molecules (about 0.07 µm), so air is no longer a continuum and the drag is smaller than Stokes’ law, as if the viscosity were η/(1 + b/pa). Without it the charges come out a few percent too large. Untick the correction in the tool and the same data give a clearly larger e.

How do I get e from a set of charges?

Every charge should be an integer multiple of e. Estimate e, divide each q by it and round to get n, then refine with e = Σqn/Σn² or the mean of q/n. The tool plots S(e) = ⟨sin²(πq/e)⟩, which has minima at e and at e/2, e/3…, and takes the largest e that fits well. Large charges make n ambiguous, so choose drops with few charges (higher balance voltage).

Which drops should I choose?

Large drops fall too fast for accurate timing and usually carry many charges; very small drops jiggle visibly from Brownian motion and fall too slowly. The usual choice is a drop that balances near 200 V and takes 10–30 s to fall 1.5 mm. Show the true values in the tool to see how large the Brownian spread is for small drops.

💧

💧Millikan Oil Drop Experiment - balance and dynamic methods, measure e

A simulated Millikan apparatus seen through the microscope eyepiece with its scale reticle; each sprayed drop has a random radius and a random number of extra electrons. As in the lab you adjust the plate voltage until a drop hovers, switch to fall and time it over a few divisions with a stopwatch, lift it back with a higher voltage, and compute its radius and charge with the balance-method or dynamic-method formula. An X-ray burst changes a drop’s charge, and after collecting many drops a histogram and a common-divisor scan reveal the elementary charge e.

Free · No sign-up · Runs in your browserLast updated: 2026-10-29

Loading tool...

How to use Millikan Oil Drop Experiment

1Press Observe, click a drop to select it and use Balance to adjust U until it hovers
2Switch to Fall and time the drop between the two yellow lines (or let auto timing do it) to get t_g
3Record with the balance method; for the dynamic method also time the rise t_e with Lift
4Pick other drops or change the charge with X-rays, or quick-collect 20 drops
5Read the charge histogram and the S(e) scan for the elementary charge and its deviation

FAQ

How does the balance method work?: First hold the drop still: qU/d = mg. Then switch the field off; drag quickly balances gravity and the drop falls at constant speed, so Stokes’ law mg = 6πηav_g gives the radius a = √(9ηv_g/2ρg) and the mass m = (4/3)πa³ρ. Combined, q = 18π/√(2ρg)·[ηl/(t_g(1 + b/pa))]^{3/2}·d/U, so only the balance voltage U and fall time t_g are needed.
How is the dynamic method different?: The drop need not be held still. A larger voltage makes it rise at constant speed and you time the rise: qU/d − mg = 6πηav_e. Together with mg = 6πηav_g for the fall this gives q = (mgd/U)(v_g + v_e)/v_g. It avoids the difficulty of setting the balance voltage exactly, at the cost of one more timing.
Why is the Cunningham correction needed?: A drop about 1 µm across is not much bigger than the mean free path of air molecules (about 0.07 µm), so air is no longer a continuum and the drag is smaller than Stokes’ law, as if the viscosity were η/(1 + b/pa). Without it the charges come out a few percent too large. Untick the correction in the tool and the same data give a clearly larger e.
How do I get e from a set of charges?: Every charge should be an integer multiple of e. Estimate e, divide each q by it and round to get n, then refine with e = Σqn/Σn² or the mean of q/n. The tool plots S(e) = ⟨sin²(πq/e)⟩, which has minima at e and at e/2, e/3…, and takes the largest e that fits well. Large charges make n ambiguous, so choose drops with few charges (higher balance voltage).
Which drops should I choose?: Large drops fall too fast for accurate timing and usually carry many charges; very small drops jiggle visibly from Brownian motion and fall too slowly. The usual choice is a drop that balances near 200 V and takes 10–30 s to fall 1.5 mm. Show the true values in the tool to see how large the Brownian spread is for small drops.

Features

Eyepiece with a 0.25 mm reticle, Stokes fall with Cunningham slip correction and true-scale Brownian motion
Balance, lift and fall voltage settings with a manual stopwatch or automatic line-crossing timer
Textbook balance and dynamic formulas, with the Cunningham correction switchable to show the systematic error
X-ray ionization changes the selected drop’s charge so its balance voltage jumps
Charge histogram and S(e) = ⟨sin²(πq/e)⟩ common-divisor scan that assigns n to each drop and finds e
Optional display of each drop’s true radius, charge and balance voltage for checking

Related tools

View all 196 Physics Tools →