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Topic 6: Fields and forces. 6.1 Gravitational force and field. Students should be aware that the masses in the force law are point masses. The force between two spherical masses whose separation is large compared to their radii is the same as if the two spheres were point masses with their - PowerPoint PPT Presentation
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Topic 6: Fields and forces
State Newton’s universal law ofgravitation.
Students should be aware that themasses in the force law are pointmasses. The force between twospherical masses whose separationis large compared to their radiiis the same as if the two sphereswere point masses with theirmasses concentrated at the centersof the spheres.
6.1 Gravitational force and field
Newton 1642-1727
Newton’s laws of gravitationAnything with mass attracts
anything else with mass. The size of that attraction is given
by my Law of Gravitation:
F = Gm1m2
r2
…where m1 and m2 are the masses of the two objects (in kg), r is the distance between them (in m) and G is “The Universal Gravitational Constant” (= 6.67 x 10-11 Nm2kg-2).
State Newton’s universal law ofgravitation.
Cavendish measurement of GClick to play
Inverse square law
What holds the planet in orbit?
Free Body Force Diagrams revision
The Earth attracts the man and the man attracts the Earth – a Newton III pair of forces where both are gravitational.
A uniform gravitational field is one where the field lines are always the same distance apart - this is almost exactly true close to the Earth's surface (Figure 1(a)).
However if we move back and look at the planet from a distance the field lines clearly radiate outwards (Figure 1(b)), getting further apart as the distance from the Earth increases.
When viewed from an even greater distance the complete field can be seen (Figure 1(c)).
Such a field is called a radial field - the field intensity (g) decreasing with distance.
Diagram 1(d) shows the distortion of the gravitational field lines by high- density rock.
Gravitational Field Strength
Consider a man on the Earth:
Man’s weight = mgBUT we know that this is equal to his gravitational attraction, so…
GMm = mg
r2
GM = g
r2
Therefore:
(this is a vector quantity)
Derive an expression for gravitational field strength at the surface of a planet, assuming that all its mass is concentrated at itscentre.
Gravitational Field StrengthDefine gravitational field strength.
Definition: Force, act, point, unit mass.
Write a definition of gravitational field strength
Determine the gravitational fielddue to one or more point masses.
Derive an expression for gravitational field strength at the surface of a planet, assuming that all its mass is concentrated at itscentre.
6.2 Electric force and fieldState that there are two types ofelectric charge.
Static Electricity
++
+ --
-
--
---
-
Conservation of charge
The law of conservation of charge states…………….
One of the fundamental laws of Physics is that charge can never be created or destroyed. Charge is always conserved in any reaction. A simple example of this is the rubbing of a polythene strip with a duster. Initially the strip and the dusted were uncharged but after rubbing the strip gains a net negative charge and the duster gains an equal amount of positive charge – the total charge in the process has been conserved.
State and apply the law ofconservation of charge.
Conductors and insulators.Describe and explain thedifference in the electricalproperties of conductors andinsulators.
In a conductor, the conduction and valence bands overlap. This allows the valence electrons to easily move along the conduction band giving the material low electrical resistance.
In insulators, there is a large forbidden energy band, which makes it difficult for valence electrons to move into the conduction band giving the material a high electrical resistance.
In semiconductors, the forbidden energy band is not too wide. Under certain conditions, electrons in the valence band can gain sufficient energy to cross the gap. This reduces the electrical resistance of the material.
The difference between conductors and insulators is………………
Coulomb’s lawState Coulomb’s law. Students should be aware that the charges
in the force law are point charges.
Coulomb’s Law
Charles Coulomb 1736-
1806
Like gravity, electrostatic force is one of the four fundamental forces. The equation
looks pretty similar too…
Coulomb’s Law
F = kQ1Q2
r2
…where k = 9.0 x 109 Nm2C-2 (the “Coulomb Law Constant”).This comes from k = 1/4πε0 …
…where ε0 = permittivity of free space (i.e. 8.85 x 10-12 Fm-1).
Electric field of a point chargeDefine electric field strength.
Write a definition of electric field strength
Students should understand theconcept of a test charge.
Determine the electric fieldstrength due to one or morepoint charges.
Electric field patterns
Draw the electric field patterns for different charge configurations.
These include the fields due to thefollowing charge configurations:a point charge, a charged sphere,two point charges, and oppositelycharged parallel plates. The latterincludes the edge effect. Studentsshould understand what is meantby radial field.
How do we predict the shape of a field?
• Imagine that you have a unit positive test charge.
• Place it in a point in the field.• Sketch the path it would take.• Repeat this many times until you have
enough field lines• The “density” of the lines represents the
strength of the field.
Point charge
+
Sphere
Positive or negative
2 Point charges
+ +
2 Point charges
+ -
Parallel plates
+
_
Edge effects
Electric FieldsElectric field strength E = F
q
(this is a bit like gravitional field strength g = F/m)
Let’s compare this to Coulomb’s Law:
Coulomb’s Law
F = kQq
r2
Putting these equations together gives us…
Electric field strength E = kQ
(in NC-1) r2
Electrostatic force and circular motionConsider an electron orbiting a nucleus in a hydrogen atom:
p
eQ. If the mass of an electron is 9.1x10-31kg and the distance to the proton is 0.11nm how fast is
the electron going?
Using mv2 = kQq we get v = 1.5x106 ms-1
r r2
A practical exampleConsider a charged polythene strip and a metal ball:
+
+ +
+--
--
- -- -
- -- -
-
Fields applet
Field for a point charge
Click to play
Electric dipoleClick to play
Falstad.com
Falstad.com
Electric fields around a point charge
Draw the edge effects for the parallel plate
2 Charged Spheres
http://www.falstad.com/emstatic/
Uniform electric fieldsConsider two charged plates:
+
-
Now consider a point charge:
QV
Work done = QV
For an electron, eV = ½mv2
Visualising the fields
Hyperlink
6.3 Magnetic force and field
Permanent magnets and domains
HyperlinkState that moving charges give rise to magnetic fields.
Draw magnetic field patterns due to currents.
These include the fields due to currents in a straight wire, a flat circular coil and a solenoid.
Field around a wire
Click to play
Field around a loopClick to play
B field for a loop
Click to play
Field patterns
Left hand Motor RuleDetermine the direction of the forceon a current-carrying conductor in amagnetic field.
Fleming's left- hand rule
Current-carrying wire in a magnetic field
S
N
F = Force
B = Magnetic
field
I = Current
Q. Where will this wire go?
Comparing magnets and solenoids
N S
Magnet:
Solenoid:
Magnetic Field around a Solenoid
Forces on a loop
Click to play
Electric motor
Click to play
Magnetic Flux DensityClearly, the size of the force on this wire depends on three
things:
1. The strength of the magnetic field
2. The current in the wire
3. The length of the wire (in the field)
These three things are related by the simple formula…
B is called “magnetic flux density” and is measured in Teslas (= 1NA-1m-1). By definition, a magnetic flux density of 1T produces a force of 1N on a 1m long wire with a current of 1A.
Determine the direction of the force ona charge moving in a magnetic field.
Hamper HL page 208 Q’s 37,38 SL Page 141 Q’s 8,9.
Force on a charged particle
Define the magnitude and direction ofa magnetic field.
A magnetic flux density of 1T produces a force of 1N on a 1m long wire with a current of 1A.
Circular pathsRecall:
++
-2 protons, 2 neutrons, therefore charge =
+2
1 electron, therefore charge = -1
Because of this charge, they will be deflected by magnetic fields:
+
These paths are circular, so Bqv = mv2/r, orr =mv
Bq
Circular paths
if angle = 90° the path is circular
if 0 < angle < 90° the path is a helix.How do you work out which bit is circular?