22 - Electric Fields

Question 1

Define electric field

Answer 1

region in space in which a charge would experience a force acting on it

Question 2

Define electric field strength + units

Answer 2

force per unit positive charge measured in NC^-1

Question 3

Electric field strength equation

Answer 3

E=F/Q
e. field strength = force/charge

Question 4

Direction of field lines for +ve point charge

Answer 4

away from centre

Question 5

Direction of field lines for -ve point charge

Answer 5

towards centre

Question 6

Direction of field assumes direction __ charge would move

Answer 6

+ve charge

Question 7

Arrows on electric field show…

Answer 7

direction of electric field

Question 8

Relationship between field lines and conductor

Answer 8

field lines at right angles (perpendicular) to conductor

Question 9

how is a stronger field shown

Answer 9

field lines closer together

Question 10

how is a uniform field shown

Answer 10

field lines are parallel and equally spaced from each other

Question 11

Point charges and uniformly charges spheres both produce…

Answer 11

radial fields

Question 12

Define one coulomb of charge

Answer 12

The amount of charge passing a point in one second when the current is 1A

Question 13

Outcome of Coulomb’s isolated charges experiment (Coulomb’s Law)

Answer 13

The force between 2 electric charges Q and q is proportional to Qq and inversely proportional to the square of their separation, r

F ∝ Qq/r^2 > F=kQq/r^2

Question 14

k constant of proportionality is equal to … therefore F =

Answer 14

k=1/4πε0
> F=Qq/4πε0r^2

Question 15

Investigating Coulomb’s law - top pan balance

Answer 15

Coat two balls in conductive paint (+ve charge) or touch each to +ve electrode oh high tension supply and charges ball lowers to create mass reading on balance. x by 9.81 to get force.

• balls are on insulating rods
• closer initial distance, higher reading, higher force

Question 16

new equation for E. field strength

Answer 16

E=F/Q and F=Qq/4πε0r^2

so > E=Q/4πε0r^2

Question 17

negative force means it is…

Answer 17

attractive force

Question 18

similarities between elec. and grav. fields

Answer 18

both produce radial fields

Question 19

differences between grav. and elec. fields

Answer 19

property that creates fields - mass vs charge
field - always attractive, direction of field always towards object vs repulsive (away) and attractive (towards) fields
diff. field stregth equations F/m vs F/Q

Question 20

elec. field between two parallel plates e.g. capacitor plates

Answer 20

is uniform - charge will experience constant force wherever it is placed between the plates

Question 21

new equation for elec. field strength (V)

Answer 21

E=V/d
units for E are Vm^-1

Question 22

capacitance on parallel plate capacitor depends on…

Answer 22

overlap area of the plates (A) and separation of plates (d)

C∝A/d
C = εA/d
ε = εrε0 εr>relative permittivity

Question 23

Millikan’s oil drop experiment (process)

Answer 23

atomsier fires oil droplets into chamber, friction in tube causes some to be charges (can use x ray source to ionise droplets), droplets fall due to grav. and some fall between plates due to top plate hole, field turned on and -ve particles attract to top +ve plate, pd adjusted for oil to be in equlibrium + stationary

Question 24

Millikan’s oil drop experiment (analysis + conclusion)

Answer 24

stationary - Fe = Fg
field off and measurements taken at terminal velocity - Fdrag = Fg

conclusion - Charge is quantised (all multiples of smallest charge recorded, e)

Question 25

motion of electron moving in direction of field

Answer 25

experience deceleration, in opposite direction experiences constant acceleration due to constant electrostatic force. attracrs +ve and repels -ve electrode

Question 26

shape of motion for particle entering field at right angle

Answer 26

parabolic path

Question 27

motion for particle entering field at right angle - horizontal motion

Answer 27

no acceleration, velocity constant
time in field > t=d/s

Question 28

motion for particle entering field at right angle - vertical motion

Answer 28

acceleration given by F=ma > a=F/m > a=EQ/m
initial velocity is 0
final velocity:
v = u + at
v = 0 + EQ/m x d/v
v = EQd/mv