Millikan's experiment

Question 1

If the charge on an electron is known then its mass can be determined from the specific charge. Describe how Millikan’s experiment with charged oil droplets enables the electronic charge to be determined. Include in your answer: • the procedures used to determine the radius of a droplet and the charge on a droplet • how the measurements made are used • how the electronic charge can be deduced.

Answer 1

1. measure the terminal speed of the falling droplet 2. at the terminal speed weight = viscous force 3. equate to give r = (9nv/2pg)^0.5 so can calculate r 4. m can be determined if r is known 5. apply pd between the plates so E=v/d and adjust until droplet is stationary 6. mg = QV/d so Q can be found 7. make a number of measurements to find Q 8. results for Q are in multiples of 1.6x10^-19C so Q can be found

Question 2

How could Milikan use his apparatus to measure the velocity v of the oil droplet and therefore the value of r

Answer 2

the microscope’s eyepiece had a graduated scale.

Question 3

Integer multiples of charge because

Answer 3

the least amount of charge is the electron of the electron = 1.6x10^-19C

Question 4

Check relativistic effects if mentions 0.9c for example

Answer 4

/

Question 5

Charge of oil droplet

Answer 5

positive

Question 6

Oil droplet is positive because

Answer 6

the electric field makes it go upwards and the electric field lines go from the positive plate to the negative plate so a positive test charge would go in the direction of the negative plate and a negative charge would go to the positive plate so as the oil droplet goes up towards the negative plate it must be positively charged.

Question 7

If QV/d > mg

Answer 7

droplet moves up due to greater electric force and reaches a terminal velocity

Question 8

Graph of speed against time for oil droplet reaching terminal velocity

Answer 8

r shape levelling off to horizontal line

Question 9

Charge calculations should always be

Answer 9

1.6*n*10^-19 C

Question 10

When an electron enters a magnetic field its speed is unchanged because

Answer 10

the magnetic force is perpendicular to the velocity so no work is done. However as the direction of the force changes the force is altered

Question 11

Electric field formed when

Answer 11

two plates, one positively charged, one negatively charged, are placed near each other

Question 12

The direction of the electric field for electron

Answer 12

Field from positive to negative which shows the direction of the electric force on a positive charge

Question 13

Increase voltage increases

Answer 13

net electric force in direction opposite to weight

Question 14

Stoke’s law

Answer 14

viscous drag force acting in the opposite direction to the velocity of a spherical object, F = 6πηrv

Question 15

Explain how he knew the charge of the oil droplet and electron

Answer 15

On being sprayed from the atomiser, the oil drop had either gained a positive charge because it had lost one or more electrons or a negative charge because it had gained one or more electrons. The oil droplet was known to be positively charged because it moved towards the negative plate when an electric field was applied meaning that it had lost electrons

Question 16

Significance of his results

Answer 16

Milikan concluded that charge can never exist in smaller quantities than 1.60x10^-19 and assumed this was the charge of the electron. Charge is quantised. It exists in packets of size than 1.60x10^-19 C – the fundamental unit of charge = size of charge carried by one electron.

Question 17

To reverse direction of force of a magnetic field

Answer 17

the magnetic flux density should be reversed

Question 18

Experiment diagram

Answer 18

Question 19

Forces diagram

Answer 19

Question 20

Electrons moving at a constant speed are directed horizontally into a uniform electric field due to two parallel plates of length L and spaced d apart which have a pd of V.

When a uniform magnetic field of flux density B is applied at right angles to the beam and the electric field the beam is undeflected. Calculate the speed of the electrons.

Answer 20

QV/d=BQv

v = V/Bd

Question 21

Electrons moving at a constant speed are directed horizontally into a uniform electric field due to two parallel plates of length L and spaced d apart which have a pd of V. At speed v

When the magnetic field is switched off, the beam is deflected vertically by the electric field by x where it leaves the field. Show that each electron takes T seconds to pass through the field and calculate the specific charge of the electron

Answer 21

t= L/v

a = 2s/t^2

F=ma=eV/d

e/m = ad/V

Question 22

Electrons moving at a speed of v are directed into a field of flux density B to form a circle of radius r. Calculate the specific charge

Answer 22

mv^2/r = Bev

e/m = v/Br

Question 23

Electrons emitted from the heated filament are accelerated through a pd of V to form a beam which is directed into a magnetic field of flux density B deflceting the beam into a circle radius r. Calculate the specific charge

Answer 23

mv^2/r = Bev

r = mv/BQ

e/m = V/Br = v^2/2vA

v = 2V/Br

e/m = 2V/B^2r^2

Question 24

A charged oil droplet of mass m is held stationary by electric field between to plates d apart. The top plate is at a positive potential of 320V relative to the bottom plate which is earthed. Calculate the charge on the droplet and state its sign and how mnany electrons are responsible for its charge.

Answer 24

QV/d = mg

e = mgd/V

+ve

e/1.6x10^-19 = number of electrons

Question 25

A charged oil droplet radius r is held stationary in an electric field of strength E which acts vertically downwards. Find the mass and the charge

density of oil = p

Answer 25

m = pV = 4/3*p*r^3

QV/d = mg

Q = mgd/V

Question 26

A charged oil droplet was held stationary betwen two plates d apart when pd was V. The terminal speed was v calculate the mass and the charge

Answer 26

6pi*n*r*v = 4/3pir^3p

r^2 = 9nv/2pg

m = 6pinrv/g

QV/d = mg

Q = mgd/V