Assessment Statements
6.1.1 state newton's universal law of gravitation
All masses in the universe attract all other masses. Every single point mass attracts every other point mass with a force that is proportional to the product of their masses and is inversely proportional to the square of their separation.
F=G(Mm)/r^2
G = universal gravitation constant (6.67 x 10-11)Nm2kg-2 determined by Henry Cavendish
M = source mass
m = test mass
r = the distance between the centers of each mass
G = universal gravitation constant (6.67 x 10-11)Nm2kg-2 determined by Henry Cavendish
M = source mass
m = test mass
r = the distance between the centers of each mass
6.1.2 define gravitational field strength
Measure of how much force a body will experience in the field. force per unit mass experienced by a small test mass in a field.
g=F/m
F = force (N)
m = mass (kg)
g=F/m
F = force (N)
m = mass (kg)
6.1.3 determine the gravitational field due to one or more point masses
Gravitational field can be shown using gravitational field lines. The gravitational field lines must be evenly spaced out around the point mass. More lines indicates a greater field magnitude.
6.1.4 derive an expression for gravitational field strength at the surface of a planet, assuming that all its mass is concentrated at its center
g=G x (M/r^2)
The equation that expresses gravitational field strength in terms of the distance away from the source must be found; a function g(r) that calculates the gravitational field strength when the distance away is known. This is because in the situation, we want to find the gravitational field strength when we know how far the surface is from the source.
The equation that expresses gravitational field strength in terms of the distance away from the source must be found; a function g(r) that calculates the gravitational field strength when the distance away is known. This is because in the situation, we want to find the gravitational field strength when we know how far the surface is from the source.
6.2.1 state that there are two types of electric charge
positive (+) = a deficiency of electrons
negative (-) = an excess of electrons
negative (-) = an excess of electrons
6.2.2 state and apply the law of conservation of change
The net charge of an isolated system is conserved so the charge stays constant. Using this law, we can always predict how many positive and negative charges exist
6.2.3 describe and explain the difference in the electrical properties of conductors and insulators
Conductors are substances that allow electrons to easily flow through them. An example is metal graphite. A superconductor is a perfect conductor and all substances become superconductors at 0 Kelvin. On the other hand, insulators are Substances that do not allow electrons to flow easily through them. Two examples are plastics and rubber and there are no perfect insulators.
6.2.4 state coulomb's law
force is proportional to the product of both charges and inversely proportional to the distance between them.
F=k x ((q1q2)/r^2))
k = constant of proportionality (9 x 10^9 Nm^2C^-2)
q1 and q2 = two charges
r = distance between two point charges
F=k x ((q1q2)/r^2))
k = constant of proportionality (9 x 10^9 Nm^2C^-2)
q1 and q2 = two charges
r = distance between two point charges
6.2.5 define electric field strength
Force per positive unit charge.
E = f/q
F= force
q = charge
E = f/q
F= force
q = charge
6.2.6 determine the electric field strength due to one or more point charges
E=k (q/r^2)
k = constant of proportionality (9 x 10^9 Nm^2C^-2)
q = charge
r = distance between two point charges
k = constant of proportionality (9 x 10^9 Nm^2C^-2)
q = charge
r = distance between two point charges
6.2.7 draw the electric field patterns for different charge configurations
Electric field always flows from positive to negative.
6.3.1 state that moving charges give rise to magnetic fields
Moving charges produce a magnetic field, according to Oersted's basic principle of electromagnetism.
6.3.2 draw magnetic field patterns due to currents
1. charge moving through wire
2. solenoid
3. freely moving charge in a magnetic field
2. solenoid
3. freely moving charge in a magnetic field
6.3.3 determine the direction of the force on a current-carrying conductor in a magnetic field
Three Right Hand Rules
1. fingers are magnetic field
2. thumb is direction of velocity
3. palm is force
Left hand is used to find the direction of the force for an electron, following the same rules.
1. fingers are magnetic field
2. thumb is direction of velocity
3. palm is force
Left hand is used to find the direction of the force for an electron, following the same rules.
6.3.4 determine the direction of the force on a charge moving in a magnetic field
Magnetic field moves from north to south and is flipped within a magnet.
6.3.5 define the magnitude and direction of a magnetic field
It is defined as a region of space where a small test magnet experiences a turning force because it rotates, not accelerates.
6.3.6 solve problems involving magnetic forces, fields and currents
1. What is the force experienced by a 40 cm long straight wire carrying a 3A current, placed in a perpendicular magnetic field of flux density 8 microT?
F = BIL
B = flux density (8 x 10^-6)
I - Current (3)
L = Length of wire (4 X 10^-2)
F = (8 x 10^-6) (3) (4x10^-2)
F = 9.7 x 10 ^-7 N)
2. A straight wire that is .7 m carried a current of 4 A in a north-south direction. If the wire is place in a magnetic field of 15 microT directed vertically downwards, what is the size of the force on the wire? What is the direction of the force on the wire?
F = BIL
B = 15 x 10^-6
I = 4
L = .7
F = 4.2 x 10^-5
Using the right hand rule, the thumb is faced downwards. The field is into the page. Therefore, the palm is facing the right, which is the East direction.
F = BIL
B = flux density (8 x 10^-6)
I - Current (3)
L = Length of wire (4 X 10^-2)
F = (8 x 10^-6) (3) (4x10^-2)
F = 9.7 x 10 ^-7 N)
2. A straight wire that is .7 m carried a current of 4 A in a north-south direction. If the wire is place in a magnetic field of 15 microT directed vertically downwards, what is the size of the force on the wire? What is the direction of the force on the wire?
F = BIL
B = 15 x 10^-6
I = 4
L = .7
F = 4.2 x 10^-5
Using the right hand rule, the thumb is faced downwards. The field is into the page. Therefore, the palm is facing the right, which is the East direction.