Chapter 16 Flashcards Preview

A2 Level physics > Chapter 16 > Flashcards

Flashcards in Chapter 16 Deck (63)
Loading flashcards...
1
Q

Electric charge

A

There is an electric field around a charged object:

-Electric charge in measured in (C) Coulombs and is either positive or negative

2
Q

Opposite charges attract and…

A

like charges repel

3
Q

If a charged object is placed in an electric field then it…

A

will experience a force.

4
Q

Electron guns make charged particles move more rapidly by…

A

sending charged particles down a potential energy ‘hill’, the higher the hill the more kinetic energy is gained. The potential energy hill is created by putting a p.d across two electrodes, positive charges then accelerate towards the negative electrode and visa versa. When going from one electrode to another, a charge loses potential energy and gains kinetic energy.

5
Q

Loss of electrical potential energy=…

A

gain of kinetic energy= mgh

6
Q

The equation for the speed of a particle

A

v= √(2qV)/m

where qV = 1/2mv^2

7
Q

Equation to calculate the force of attraction or repulsion between two point charges…

A

F= (K.Q1.Q2)/(R^2)

Where k = 1/ 4πε

8
Q

The force on Q1 is always…

A

equal and opposite force on Q2

9
Q

The further apart the chares are…

A

The weaker the force between them

10
Q

Electric field strength

A

E= force per unit positive charge, the force that a charge of +1C would experience when placed in the electric field.

11
Q

all particles with the same charge gain the same kinetic energy when they move though the same potential difference, but…

A

Some may move faster than others depending on their size

e.g. a electron moves much faster then a proton because its mass is much smaller

12
Q

Equation for electric field strength

A

E= F/q

Where E = electric field strength (NC^-1), F= force (N) and q= charge (C)

13
Q

Gravitational field strength

A

g= F/m

where g= gravitational field strength (NKg^-1), F= force (N) and m = mass (kg)

14
Q

Radial field

A

-In a radial field, E depends on the distance r from the point charge Q.

15
Q

Equation for E in a radial field

A

E = kQ/ r^2

E is inversely proportional to r^2

16
Q

The further you go away from Q…

A

The field strength decreases and the field lines get further apart.

17
Q

Electrical potential energy

A

-The work that would need to be done to move a small chare, q, from infinity to a distance r away from a point charge Q.

18
Q

Equation for energy in an electric field

A

E= KQq/ r

19
Q

Repulsive force field

A

Graph: E decreases as r increases
-In a repulsive force field ( positive charge, lines away from Q)- you have to do work against the repulsion to bring q closer to Q. The charge q gains potential energy as r decreases.

20
Q

Attractive force field

A

Graph: E increases negatively as r increases positively

  • In an attractive field, the charge q gains potential energy as r increases.
  • Also gradient of a tangent gives he electric force at that point.
21
Q

Electric potential..

A

V= electric potential energy per unit positive charge

22
Q

Equation for electric potential energy

A

V= E(electric)/ Q and substituting for E gives V = KQ/r

23
Q

V is measured in…

A

Volts or joules per coulomb

24
Q

As with E, V is..

A

Positive when the force is repulsive and negative when the force if attractive (same graphs)
-Also the gradient f a tangent gives the field strength

25
Q

Field strength is the same everywhere in…

A

A uniform field

26
Q

Uniform field can be created using…

A

Two parallel plates to the opposite poles of a battery

27
Q

Between these plates

A

The field strength is the same at all point

28
Q

Equation for field strength in a uniform field

A

E= v/d where v= the p.d and d= the distance between the plates

29
Q

E can be measured in …

A

Vm^-1

30
Q

The field lines are

A

parallel to each other

31
Q

The surfaces are

A

Equipotential and are parallel to the plates, and at 90 degrees to the field lines.

32
Q

Principle of linear acceleration

A

-The negative electron will accelerate to the positive electrode

33
Q

Switching P.D to keep accelerating electrons

A
  • The first group of negative electrons will accelerate towards the positive electrode.
  • The p.d is then alternated so once they pass the positive electrode, it is now negative, so the electron accelerate to the positive electrode in-front of it.
  • The alternating p.d switches back and forth so that the electrons are accelerated as they pass between successive electrodes.
34
Q

If the electron is accelerating, gaining potential energy, the change in the potential Δ V and the change in potential energy qΔV must both be…

A

Negative!

Change in kinetic energy = -change in potential energy

35
Q

The change in the kinetic energy is produced by a force, F….

A

change in energy = FΔx = -qΔV

Force on the particle =

F= - qΔV/ Δx

Force = - potential energy gradient

36
Q

Therefore Force on particle can be arranged to…

A

E= F/ Q => - ΔV/ Δx

electric field strength = -potential gradient

37
Q

To find the uniform electric field between a pair of conducting charged plates…

A
  • Put a voltmeter across to measure the p.d (V)
  • Measure the distance between plates (d)
  • The field is uniform, so the equipotentials are equally spaced and the gradient is the same right across the gap.

=» E = V / d

Units = Vm ^-1

38
Q

Similar units…

A

NC^-1 = Vm^-1

Nkg^-1 = ms^-2

39
Q

Field lines are always … to eqipotential surfaces

A

Perpendicular

40
Q

Drawing field lines

A
  • Always start and end on charges
  • Cannot cross each other
  • Always perpendicular to equipotentials
  • Point from higher potential (more positive) to lower potential
  • Close together in strong fields, far apart in weak fields
41
Q

Millikans experiement

A
  • Used an atomiser to fire a mist of ionised oil drops that are charged by friction s they leave the atomiser
  • Some fell through the hole and were viewed in the microscope
  • He then applied p.d across the plates to produce a uniform field that exerts a force on the charged drops. By adjusting the p.d he varied the strength of the field.
  • He applied enough p.d until the drops were stationary so when Upwards force = Downwards force
  • The voltage at which the oil was stationary was measured, the charge was then calculated
42
Q

When the oil drop is held stationary

A

F= qE and W= mg

=> qE = mg
=> E= V/d
=> V= mgd/q

=> V = mgd/ ne
(If the charges are multiples n of electron charge e, the nq= ne)

43
Q

Conclusion to Millikans experiment

A
  • Charge exists in discrete ‘packet’ size 1.6 x 10 ^-19 C.
  • He found that the charges were all multiples of 1.6 x 10^-19, thus showing that each drop was made up of smaller charges with a charge of 1.6 x 10^-19 (electrons)
44
Q

Charged particle are affected by magnetic fields and so a current-carrying wire…

A

Experiences a force in a magnetic field

45
Q

Equation for the force exerted on a wire in a magnetic field perpendicular to the field

A

F = BIL

46
Q

A charged particle moves a distance l in t…

A

Velocity= L/ t
and I = q/t

so t = L/ v

47
Q

I= qv/ L goes in F = BIL so …

A

F= qvB

48
Q

By using flemmings left hand rule…

A

The force is always perpendicular to the magnetic field

49
Q

Centripetal force

A

The centripetal force and the electromagnetic force are equivalent for a charged particle along a circular path

50
Q

If F = mv^2 / r where f=qvB then…

A

qvB= mv^2 / r

So r = mv/ Bq

51
Q

When a electron beams are deflected …

A

There will be a positive plate above or below the deflection path to make the electron no stay in a straight line

52
Q

If the electron beam is staying stationary with a positive plate above and a negative plate below, when the electric field is turned on, what is happening?

A

The charge is must be negative as It wants to travel up as it is attracted to the positive plate, however the force of gravity is pulling I down so it remains stationary

53
Q

The momentum of a particle is proportional to…

A

The radius of the curvature of the path
Therefore p = qrB

where p = momentum

54
Q

Disadvantage to linear accelerators and advantage to circular accelerators

A

D- They have to be extremely long in order for the particle to reach a high energy

A- Particles with the same mass but different charges can be accelerated in opposite directions

55
Q

The smaller the rings are in circular accelerators…

A

-the larger the centripetal acceleration a so it takes away energy from the particles

56
Q

Charged spheres

A

-Small particles behave like small charged spheres

57
Q

The electrical field of a sphere obeys the inverse square law

A

E = k q / r^2

58
Q

The electric field strength near an isolated charged sphere obeys an inverse square law

A

E ∝ 1/ r^2

59
Q

The field strength must also depend on the amount of charge therefore…

A

E ∝ q/ r^2

60
Q

FIELD STRENGTH

A

E = -K Q / r^2

61
Q

FORCE

A

Field strength x q

E = -K Q q/ r^2

62
Q

ENERGY

A

Force integrated

We = K Q q/ r

63
Q

POTENIAL

A

Energy ÷ q

Ve = K Q/ r