FIELDS
Electrical field
+ve charge moving towards –ve charge
Higher potential
+
V
x
V
Gradient of slope =
x
Lower Potential
Increases if V bigger
and x smaller
- Potential gradient = Electrical Field Strength
Field Strength Line patterns
(line to indicate direction of force felt by a +ve charge)
-
-
+
+
+
Non-uniform fields
Uniform field
Lines of equipotential – no work done on charge
Field strength
lines
Equipotentials in
a uniform field
will be equally
spaced
Equipotential,
90 to field
strength lines
800V
0V
200V
400V
600V
Describing a uniform field
Electric Force
• Force between two charges
• May be attractive (-ve) or repulsive (+ve)
Symbol = F
Units = N
Describing a uniform field
Electric Field Strength
• Force per unit +ve charge
Important due to
repulsive/attractive
nature of force
Symbol = E
Unit = N
-1
C or
E=F
q
V
-1
m
E=V
d
distance between
charges (plates)
Describing a uniform field
Electric Potential Energy
• Energy of a +ve charged particle (q)
due its position in an electric field
Symbol = EPE
Units = C V = J
EPE = qV
Acceleration due to electric field
EPE = qV
+
+
10V
0V
EPE = 0
 potential energy =  kinetic energy
qV = ½mv2
Important point:
K.E. gained is the same for both particles
Velocity is different due to difference in mass
Electron velocity quickly approaches the speed
of light - relativistic.
Velocity equation only valid when v << c
The electron volt (eV)
A measure of energy
qV =
2
½mv
1 eV = Energy gained when an electron is
subjected to a potential difference of 1V
1 eV = 1.6 x 10-19C x 1V = 1.6 x 10-19 J
Millikan’s Oil Drop Experiment
to measure charge on an electron
Forces experienced
by oil drop?
+
F (electric attraction) = qE
F (weight) = mg
Oil drop
(charge q)
-
If the oil drop hovers then…
mg = qE
The cunning bit...
Add or remove electrons
Then ionise oil drop (using radioactive source)
Re-adjust the voltage (and therefore the
size of E) to make oil drop hover again.
Change in voltage proportional to charge on
an electron
Kepler’s Laws
Kepler: Geometry rules the Universe
Law 1: a planet moves in an ellipse with the Sun at one focus
Astronomy
Geometry
planet
Mars
a
b
Sun
focus
focus
Ellipse: curve such that sum of a and b is constant
Orbit of Mars an ellipse with Sun at a focus
Kepler 2
Kepler: Geometry rules the Universe
Law 2: the line from the Sun to a planet sweeps out equal areas in equal times
Astronomy
Geometry
planet
Mars
fast
slow
focus
Sun
Speed of planet large near Sun, smaller away from Sun
Areas swept out in same time are equal
Kepler 3
Kepler: Geometry rules the Universe
Law 3: square of orbital time is proportional to cube of orbital radius
2
4
Mars
Orbital period against
orbital radius
Orbital period squared
against orbital radius
cubed
Mars
3
1
2
Earth
Venus
1
Earth
Venus
Mercury
Mercury
0
0
0
50
100
150
radius/million km
200
250
0
1
2
radius3/AU3
3
4
Apple and moon experiment
by Newton
Question - how does gravity extend into space?
Newton calculated...
Acceleration of Moon towards Earth
Acceleration of an apple towards Earth
Conclusion...
Angular acceleration of
Strength of
moon towards earth
= earth’s gravity
(due to circular motion)
at 60RE
Gravity changes at a rate of inverse
square of distance
This extended gravitational force out
into the universe - an amazing result (!)
Gravitational force, F
F = -Gm1m2
2
r
F = gravitational force, N
G = gravitational constant, 6.67 x 10-11 N m2 kg-2
m1 = mass of first object, kg
m2 = mass of second object, kg
r = distance between the two objects, m
Earth orbiting the sun
This is the
gravitational
force between
the sun and
earth
Requirement for an object to orbit
Fgravitational = Fcentripetal
90 to velocity
of object
Needed for
circular motion
Gm1m2 = m2v2
2
r
r
N.B. m1 = mass at centre of orbit, m2 = mass of satellite
Remembering v = 2r
Very important
T
Gm1m2 = m242r
r2
T2
2
T
=
2
3
4 r
Gm1
(Kepler III)
The satellite must be travelling fast
enough for its orbit radius (Kepler III)
• Not faster enough - orbit will collapse
• Too fast - will overcome gravitational
forces and escape
Types of orbit
Geostationary Relative to the earth it
doesn’t move
T = 24 hours
Polar
Orbits N-S (over the poles),
the earth rotates and so it
looks at a different place each
orbit.
T = 90 minutes
Gravitational field strength, g
- better known as gravity
If there is a force there is an acceleration
F = ma
If the force is due to gravitational forces then
acceleration is acceleration due to gravity
F = mg
Or
g=F
m
In words, gravitational force per unit mass
acting at a point
Know
F = -GMm
r2
Then
g = -GMm = -GM
mr2
r2
units N kg-1
Uniform Field (near surface of Earth)
Gravitational Potential Energy
= mgh
Stored ability to do work - something else
has done work to get the object to that point
Remember:
work = force x distance (in direction of force)
= mg x h
Change in GPE
mg(h + 3Δh)
mg(h + 2Δh)
mg(h + Δh)
mgh
Δh
h
h=0
Work done moving an object by Δh
Force (mg) /N
(near the earth’s surface, g  constant)
Area = mg Δh
= work done
Δh
h1
h2 Height /m
Gravitational potential
...“The work done to move unit mass
from infinity to that point”...
Symbol = V
V = mgh = gh
m
Unit = J kg-1
Don’t forget
hr
Gravitational potential is the total work,
against the gravitational force, for 1kg
to go from a point where g = 0 to the
point in question where g = x N kg-1.
g = 0 N kg-1 at r = 
g = 9.8 N kg-1 at r = 6.4 x 106m
A convention...
Earth only
calculations
Space
calculations
GPE = x
GPE = 0
GPE = 0
GPE = - x
Force (mg) / N
Work done moving an object
from  to r (Δr)
r /m
Δr

0
r
Area = work done


= m g dr
r
N.B. g isn’t constant (non-uniform field)
Gravitational potential - work done on
unit mass i.e. m = 1kg


V = g dr

r
V = - GM dr =
r2
r

GM
r
r
= - GM
r
Another interesting point...


V = g dr
r
Can be rearranged to ...
g = - dV
dr
The gradient of gravitational potential is
gravitational field strength.
Gravitational force
F = - GMm
2
r
Gravitational
field strength
g = - GM
2
r
Gravitational
potential
V = - GM
r
Energy conservation
Etotal = Ekinetic + Epotential
mgh = Etotal - ½mv2
gh = constant - ½v2
V
2
-½v
r
-½v2
r
V
The gradient of either graph is g
Escape velocity
V = 0, r = 
V = -62.5MJ kg-1
Energy gained if falling into hole
Energy required to get out of hole
Stationary object (v = 0),
at V = 0
½mv2 + mV = ETotal

0+0=0
Nudge object into well, ETotal = 0
K.E. increases as P.E. become more -ve
½mv2 + mV = 0
v   -2V
V is -ve
At Earth’s surface V = -62.5MJ kg-1 , a
1kg mass will hit the ground at ~11km s-1
if nudged into well.
Conversely...
A 1kg mass launched at 11km s-1 will just
make it to V = 0, the brim of the potential
well.
11km s-1 = escape velocity