Millikan's Oil Drop Experiment Flashcards
what was the purpose of the oil drop experiment
to determine the charge of an electron
what was the basic setup for the experiment
- everything happened within a chamber with a viewing microscope
- a cathode and anode is fixed inside the chamber
- with a vaporiser attached to the top end of it
what is the purpose of the viewing microscope attached to the chamber
- to closely observe the oil drops
- specifically their movments
what else would therefore have to be attached to the closed off chamber for the microscope to be of any use
a light source
where would the microscope and light source be connected to relative to the cathode and anode
- in between them
- the light source cant be aligned with the microscope however
what is the purpose of the cathode and anode
to create a uniform electric field within a region of the chamber
between the cathode and anode, which ones would be at the top (technically in the middle of the chamber) and below (at the bottom of the chamber), as well as what type of pd would be required
- the anode (+ve) would be at the top
- while the cathode (-ve) would be at the bottom
- a variable pd would be needed
what is different between the anode and the cathode (other than their charge) and why
- the anode has a small gap in the middle of it
- for the oil drops to fall through and enter the electric field
what is the purpose of the vaporiser
to vaporise the oil as it is sprayed into the chamber
what was the basic premise of the experiment
- oil droplets were sprayed into the chamber, in which some would gain a negative charge due to this
- as the drops fall due to gravity, some will enter the anode’s gap and enter the field
- one drop would be focused on and have the pd of the electric field adjusted until the drop was stationary within it
why did some oil drops gain a negative charge
- because of the friction between them and the tube theyre squirted out of
- because of the transfer of electrons due to friction, some drops would have some multiple of negative charge (relative to an electron)
- while some would actually be positively charged due to electron removal
what is a more scientific method of giving the oil drops a charge now-a-days
x-ray ionising radiation
when the negatively charged oil drops entered the electric field, what direction is the field exerting a force on the drop and why
- in the opposite direction to its weight (upwards)
- because the positively charged plate is at the top
- so the negatively charged drop would be attracted to the positive terminal above it
how could that be deciphered from the uniform electric fields field lines
- the field lines point in the direction that a positive charge would have a force exerted on it if were in the field
- therefore the field lines would point downwards (+ve to -ve)
- as the oil drop is -vely charged it would simply have a force exerted on it in the opposite direction, upwards
therefore, why would you need to be constantly adjusting the variable pd when looking at an oil drop through the microscope
- you would firstly need to increase the pd to counteract the drops acceleration due to gravity
- this likely means the force due to the electric field strength is larger than its weight
- therefore, when the oil drop begins moving upwards, you need to reduce the pd
- this process would be repeated until the drop remains stationary
what is the relationship between the force from the field and the weight of the oil drop when it is stationary
weight = electric force
although having the oil drop stationary is the best indicator for balanced forces, why could the weight equaling the force from the field still result in movement of the drop
- because weight equaling the force from the field means there is no acceleration, not velocity
- so the drop moving at a constant velocity also means they equal each other
- although this is simply harder to verify
a bit late now, but why does increasing the pd increase the force exerted on the droplet by the field
- because electric field strength increases (E = V / d)
- meaning force increases (F = EQ)
what thought process did Millikan have when calculating the weight of the oil drop
- the density of air so is low so upthrust is negligible
- this means that when the drop reaches terminal velocity, viscous drag is equal to its weight
what equation can be derived from this
mg = 6 pi r n v(terminal) from stoke’s law
what is n and r
- n is the viscosity of air
- r is the radius of a drop
what equation is derived from the fact that weight = electric force when the drop is stationary
- mg = EQ
- as E = V / d
- mg = VQ / d
how would you rearrange for Q using those two equations
- by equation them together
- VQ / d = 6 pi r n v
- Q = 6 pi r n v d / V
because millikan couldnt measure the radius of the droplets, how did he remove the variable r from the equation
- by rearranging another one of stokes law
- v(term) = (2r^2 g (p(oil) - p(air))) / 9n
- into r = the root of (9nv(term) / 2gp(oil))
so what was the final formula he came up with
Q = (6 pi n v d / V) * (9nv / 2gp(oli))^1/2
after repeating the experiment hundreds of times, which did millikan discover
- the charges of the droplets were always a multiple of 1.59x10-19
- which was less than 1% away from the 1.602x10-19 value we use today
what scenario would adjusting the pd be alot easier
- when the droplet is already at terminal velocity
- as it is no longer accelerating due to gravity
- so the steady increase of pd would slowly be decelerating it until it became stationary
