Oil droplet could be determined. Repeating the experiment for many droplets, he found that the values measured were always multiples of the same number. He interpreted this as the charge on a single electron: 1.6 × 10−19 coulombs. Let’s look at the experiment in detail.
Initially the oil drops are allowed to fall between the plates with the electric field turned off. They very quickly reach a terminal velocity at which point the gravitational force is balanced by the frictional force due to air resistance (blue arrow balances red arrow in figure). A single oil droplet is selected and its terminal velocity v1 is measured. The drag force acting on the
drop can be worked out using Stokes' law:
where v1 is the terminal velocity of the falling drop, η is the viscosity of the air, and r is the radius of the drop.
The volume of the droplet is simply and the density is and so the mass of the droplet is
and the weight is . Since when the droplet is at terminal velocity we
can write:
and by rearranging we can say: Once r is calculated, W can easily be determined.
Now the electric field is turned on and the droplet experiences a force due to the electric field of where q is the charge on the oil droplet and E is the electric field between the plates. The
electric field is then adjusted until the oil droplet remains steady, meaning that the force due to the electric
field is equal to the weight of the droplet. So and therefore
Since the weight of the droplet has already been found, the charge can be calculated.
Doing this for a huge number of droplets he discovered that the values calculated for the charge q on each droplet were all integer multiples of -1.60×10-19 C. He therefore stated that this was the charge of the electron and that the droplets contained multiple numbers of electrons, i. e. q = n×(electron charge) with n an integer number of electrons.
Using Thomson’s result for the charge/mass ratio, the electron mass was then me = 9.1×10-31 Kg. 1.6 Electron-volts
This is such a tiny topic but it is also one of the most important ones. You will use electron-volts in two main ways; (i) as a convenient unit of energy, and (ii) to calculate the velocity of a charged particle passing through an electric field.
Definition: The change in energy E of a charged particle q as it moves through a potential difference V is given by . 1 electron-volt =1eV =1.6×10-19×1 = 1.6×10-19 J.
(i) In particle physics we often deal with very small energies and it is often therefore more convenient to refer to 3.2×10-19 J as 2 eV for example.
(ii) Imagine a particle of mass m and charge q is accelerated from rest through a voltage V. The energy given to the electron is therefore qV . If all this energy is converted into motion then this will be equal to
the final kinetic energy of the particle. So and the velocity of the electron is where v is in metres per second, V is in volts, q is in coulombs, and m is in kilograms?
.