Physics Beyond 2000

Chapter 13 Electrostatics

http://www.sciencejoywagon.com/physicszone/lesson/07elecst/default.htm

Electrostatics

• The charges are at rest. There is not any electrical current.

• The charges have forces on each other.

A neutral atom• Each electron carries one unit of negative

charge.

• One unit of negative charge = -1.6 × 10-19C

A neutral atom• The nucleus contains protons.

• Each proton carries one unit of positive charge.

• One unit of positive charge = 1.6 × 10-19C

A neutral atom• For a neutral atom, the number of electrons

= the number of protons.

• Total charge of a neutral atom = 0

A charged atom

• If a neutral atom gains electrons, it becomes negatively charged.

A charged atom

• If a neutral atom loses electrons, it becomes positively charged.

A charged atom

• If a neutral atom loses electrons, it becomes positively charged.

Charges on insulators

• Excessive charges on insulators will stay on the insulator.

• Example: a plastic ruler with positive charges on it.

++ +++ +

Charges on insulated conductor

• The excess charges on the insulated conductor remains constant unless the conductor touches another conducting medium.

conductor

insulated handle

• When a conducting medium touches the conductor, the excess charges flow away.

conductor

insulated handle

conducting rod

• If all the charges flow away, the conductor becomes neutral.

• This is called earthing.

conductor

insulated handle

conducting rod

• In some cases, only part of the charges flow away. Some charges still remain.

• This is called sharing.

conductor

insulated handle

conducting rod

Charges on isolated conductor

• The conductor is completely isolated from any other object. Its excess charges remain constant.

Two kinds of charges

• Positive charges

• Negative charges

• Unit: coulomb ,C.

• One proton contains 1.6 × 10-19 C.

• One electron contains - 1.6 × 10-19 C.

• Like charges repel.

+ ++ ++ +++

• Like charges repel.

- -- -- -- -

• Unlike charges attract.

- +- +- +- +

http://www.colorado.edu/physics/2000/waves_particles/wavpart2.html

Charging objects

• Use a power supply

• By rubbing

• By induction

Charging by power supply• Use a light conducting sphere and a EHT.

• Ground the negative terminal of the EHT.

• Let the sphere touch the positive terminal of the EHT.

• What kind of charges is on the sphere?

Positivecharge

earthed

Charging by power supply• Use a light conducting sphere and a EHT.

• Ground the positive terminal of the EHT.

• Let the sphere touch the negative terminal of the EHT.

• What kind of charges is on the sphere?

Negative charges

earthed

Charging by power supply

• Explain why the sphere is charged.

Hint: sharing the charges.

Charging by rubbing• Use a piece of dry cloth to rub a polythene

• What kind of charges is on the polythene rod?

Negative charges

Charging by rubbing• Explain why the polythene rod is negatively

charged

Shuttling ball experiment

positivenegative

1. Metal plate B connected to the positive terminalof the EHT.

2. Metal plate A connected tothe negative terminalof the EHT.

3. An isolated light conducting sphere.

1. Connect tothe positive terminalof the EHT.

2. Connect tothe negative terminalof the EHT.

4. Allow the ballto touch one plate

1. Connect tothe positive terminalof the EHT.

2. Connect tothe negative terminalof the EHT.

5. The ball isshuttling.

1. Metal plate connected to the positive terminalof the EHT.

2. Metal plate connected tothe negative terminalof the EHT.

Explain the experiment.

Charge distribution in a conductor

• Inside the conductor: the excess charges do not stay inside the conductor because of the repulsion.

• On the surface of a conductor: the excess charges reside on the surface of the conductor. +

++ + +

+ + + + + + + + +

Charge distribution on the surface of a conductor

• If the surface of the conductor is flat, the charge distribution is uniform except at the edge.

• If the surface of the conductor is spherical, the charge distribution is uniform.

++ + +

• If the surface of the conductor is of pear-shaped, the charge distribution is dense on curved surface than on flat surface.

+ + +++++

• On the surface of a conductor, the surface charge density describes the distribution of charges.

• Surface charge density is the charge per unit surface.

++ + +

Q where Q is the amount of charge on an area A.

+ + + + + + + + +

Example 2

• The surface area of a sphere is 4r2.

• The SI unit of charge Q is coulomb (C)

• The SI unit of surface charge density is

coulomb per m2 (C m-2) .

Electric fields

• There are forces (attraction or repulsion) between charges.

• How to describe these forces?

(How large is the force and what is its direction if I put a charge, say 1 C, at a point?)

• Electric field is the basic concept.http://www.colorado.edu/physics/2000/applets/nforcefield.html

http://www.colorado.edu/physics/2000/waves_particles/wavpart3.html

Electric fields

• An electric field is a region in which an electric charge experiences a force.

Place a charge at this point. There isnot any force on it. So the electricfield is zero at this point.

Electric fields

• An electric field is a region in which an electric charge experiences a force.

Place a charge at this point. There isforce on it. So there is an electricfield here.

http://www.sciencejoywagon.com/physicszone/lesson/07elecst/fieldint/efield.htm

Electric fields

• There are two ways to describe the electric fields.

(i) Lines of forces (graphical) ;

(ii) Electric field strength (numerical).

Lines of Force• Draw directed lines.

• Some electric field patterns:http://www.sciencejoywagon.com/physicszone/lesson/07elecst/static/fieldmap.htm

Lines of Force

• A field line always directs away from a positive charge and ends at a negative charge.

• The tangent to the line at any point gives the direction of force acting on any positive test charge.

• The number of field lines drawn per unit cross-sectional area is proportional to the strength of the electric field.

Lines of Force

• A field line always directs away from a positive charge and ends at a negative charge.

Lines of Force

• The force is in opposite direction if the test charge is negative.

Lines of Force

Lines of Force• The number of field lines drawn per unit

cross-sectional area is proportional to the strength of the electric field.

2. Strong electric field

1. Weak electric field

Electric field strength E

• The electric field strength E at a point is the force per unit test charge placed at that point.

FE where F is the force on test charge q.

Unit of E: N C-1 or V m-1

q = 3 C

F = 18 N

• ExampleFind the electric field strength at the point of the charge.

Electric field strength E

• Note that the test charge q must be small.

• For a big test charge, the original electric field may be changed.

1. The electric field of a single charge

2. The electric field changes dueto the presence of another big charge.

Ionization of air by strong electric field

1. An air molecule consist of equal numberof positive and negative charges.

2. In a strong electric field, positive charges arepulled to one side and negative charges to theopposite side.E

E3. Charges are completely separated. Thereforms an ion-pair.

forces

Ionization of air by strong electric field

• Sharp point carrying charges may create a very strong electric field.

• The electric field strength that causes ionization is called the breakdown field strength E

b. For air, Eb 106 NC-1.

+ + + + + + ++

+ + + + + + +

Example 3

FE F = q.E•

• Electric force is large in comparison with gravity.•Effect of gravity is usually ignored when electric force is present.

Uniform Electric field

• In a uniform electric field, the electric field strength E is the same at every point.

• Force on a charge in a uniform electric field is constant at every point.

F = E.q = constant

• The field lines (or lines of force) are evenly spaced and parallel for a uniform electric field.

• Making a uniform electric field:

1. Two parallel metal plateswith one connected to highpotential.

High potential

2. Another metal plate is connected to low potential.

Low potential

3. Between the plates, there is a uniformelectric field

• Making a uniform electric field:

High potential Low potential

4. The lines of force are from the plate of high potentialto the plate of low potential.

• Checking a uniform electric field:

High potential Low potential

5. Use a charged metal foil to test the electric field.6. Move the foil from left to right. The angle of deflection of thefoil should not change, indicating a constant electric force.

Insulatinghandle

Charged metal foil

• Forces on the charged foil are in equilibrium.

3. electric force onthe charges

2. tension along the foil

4. weight of the foil

foilcharged foil

1.Angle ofdeflection

verticalline

Uniform Electric field• Forces on the charged foil are in equilibrium.

• Find FE in terms of W and .

FE = W.tan

Example 4

• 50 kV m-1 = 50,000 N C-1

• 2 C = 2 10-6 C

• The gravity is ignored in this example because it is too small.

Motion of a charge in a uniform E-field

1. A positive charge q is projectedwith initial velocity u, making anangle with a uniform electric field.2. The electric field strength is E. 3. The path of the charge will be aparabola.4. Prove it.5. Hint: Separate the motion intotwo, one along horizontal direction and the other vertical. x

Work done by the field

• Suppose that the charge q moves freely in a uniform electric field E for a displacement d.

• What is its gain in kinetic energy?

• What is the work done by the field force?

q EW = Eqd

Coulomb’s Law• Numerical method to find the force between

two charges – Coulomb’s law.

• For two point charges Q1 and Q2 at distance

r apart, the electric force between them is

Coulomb’s Law• For two point charges Q1 and Q2 at distance

r apart, the electric force between them isQ1

where is the absolute permittivity of the intervening medium.

Coulomb’s Law

1. The absolute permittivity of free air is o = 8.9 10-12 C2 N-1 m-2

2. For other medium, its absolute permittivity is expressed in terms of the relative permittivity r and the absolute permittivity of free air o by

= r . o .Refer to table 13-1 on page 282.

3. For free space,

129100.94

1 CNmo

Investigate Coulomb’s law

1. A suspended charged sphere X of mass m.2. The length of the nylonthread is L.

nylon thread

2. Put another charged sphere Y nearto X.3. There is an angle of deflection .4. Measure the separation r and dthe displacement of X .

Objective: To show that2

1. Find the electric force in terms of m, d and L.2. Use small angle of deflection tan sin 3. This shows that F d.

1. Repeat the experiment by changingthe separation r. The charges should notchange.2. Plot a graph of d against

http://www.engr.uky.edu/~gedney/courses/ee468/expmnt/coulomb.html

Electric forces

• The electric forces act along the line joining the two point charges.

• The electric forces may be added as vectors.

Analogy to gravitational force

Coulomb’s law Law of gravity

Inverse square law F

Attraction or

repulsion

Attraction only

Negative force for attraction and positive force for repulsion

Analogy to gravitational force

• http://www.pa.msu.edu/courses/1997spring/PHY232/lectures/coulombslaw/grav%20_analogy.htm

mass 1 mass 2

r Gravitational forceis always attraction.

positive charge 1 negative charge 2

r Electric force may berepulsion or attraction.

Example 5

• An electron contains one unit of negative charge Q1 = e = -1.6 10-19 C.

• An -particle has two units of positive charge Q2 = + 2 1.6 10-19 C

• The electric force is an attractive force. So its value is negative.

Shell Theorem• A uniform spherical shell of charge behave,

for external points, as if all the charges were concentrated at its centre.

1. A metal sphere withcharge Q1 uniformlydistributed.

2. Separation = r>R 3. A point charge Q2

4. The force on Q2 is

Shell Theorem• A uniform spherical shell of charge exerts

no force on a charged particle placed inside a shell.

1. A metal sphere of radius Rwithcharge Q1 uniformly distributed on its surface.

2. Separation = r< R

3. A point charge Q2

4. The force on Q2 is zero.

Shell Theorem• The force between two spherical distribution

charge is equal to that of two point charges.

++ + +

rQ1 Q2

3. What is the magnitude of F?

1. Spheres

2. Point charges

Calculation of Electric field strength (E)

• E due to a point charge +Q

The electric field strength at adistance r from a point charge +Q is

The field is spherically symmetrical.

E = ?q

Example 6

• Is an alpha-particle a point charge?

• What is the quantity of charge of an alpha-particle?

• What is the distance 1 m?

• The unit of E is either NC-1 or Vm-1.

• E due to a charged spherical metal shell.

1. The charges reside on the surface of the spherical shell.

2. The E inside the shell is 0.

3. The E outside the shell is similar to that of a point charge at the centre of the shell.

1. E = 0 inside the conducting shell.

2. Outside the shell

3. What is E juston the surface ofthe shell?

Q3. What is E juston the surface ofthe shell?

4. Use the surfacecharge density to simplify the answer.

• E due to a charged spherical metal shell.E

1. E = 0 inside the conducting shell.

3. Outside the shell

RE2. On the surface

• E on the surface of any charged conductor

Note the may be different on different positions of the non-spherical conductor. is small on a flat surface and high on curved surface.

• E on the surfaces and between two infinite large and parallel charged sheets

E1. Positivelycharged plate 2. Positively

charged plate

3. Constant electricfield strength between

Example 7

• The medium between the parallel plates is air.

• Use

E where o = 8.9 10-12 C2 N-1 m-2

• E due to a uniform spherical charge.

+ + + + ++ + + + ++ ++ + + + ++ + + + ++

+ + + + +

+ + A metal shellor metal sphere

1. charges are uniformlydistributed in the sphere

2. Charges are on the surface only.

+ + + + ++ + + + ++ ++ + + + ++ + + + ++

+ + + + +

1. Total number of charges = Qo

2. Outside the sphere at a distance r from the centrer

3. Inside the sphere at a distance rfrom the centre

rQE or

Radius = R

+ + + + ++ + + + ++ ++ + + + ++ + + + ++

+ + + + +

rQE or

QE or 1.

Gauss’ Law

• Gauss’ law is a convenient way to calculate the electric field strength E, especially when Coulomb’s law cannot be applied.

Gauss’ law

• Definition of Electric flux :

The electric flux across a surface area isthe product of the surface area A and the normal component of the electric fieldstrength E

= E . A

Example 8

• Note that we have to calculate the E in the second part.

Find the E and thus from the right diagram.

Gauss’ law• The total electric flux normal to a closed

imaginary surface is equal to

where Q is the charge enclosed by an area and is the permittivity

1. Suppose thatthere is a chargeQ.

2. Draw an imaginary surface to enclose the charge

3. The electric fluxthrough this surfaceis always

Gauss’ law• Use Gauss’ law to find the electric field strength

at a distance r from a point charge Q.

1. Draw a spherical surfaceof radius r around the charge Q.2. What is the total electric

flux?3. Find E from = E .A. Note

that E is the surface.

Gauss’ law• Use Gauss’ law to find the electric field strength at a distance

r from a line of charge with charge per unit length .• Note that it is difficult to use Coulomb’s law here.

rxEr = ?

Gauss’ law• Use Gauss’ law to find the electric field strength at a

distance r from a line of charge with charge per unit length .

rxEr = ?

1. Draw a cylindrical surfaceof radius r and height h around the

line of charge.2. What is the total electric flux?3. Find E from = E .A. Note

Gauss’ law• Find the electric field strength E due to a

hollow charged conductor enclosing a point charge.

Gauss’ law• In the cavity of the hollow sphere with

0 < r b.

1. Draw a spherical surfaceof radius r around the charge Q1.2. What is the total electric

Gauss’ law• Inside the conducting sphere with b r a.

+Q2+Q1

1. Draw a spherical surface of radius r inside the conducting sphere.

2. What is the charge enclosed by this spherical surface?

3. What is the total electric flux?4. Find E from = E .A. Note that

E is the surface.+Q1

E2 = 0-Q1 +Q1+Q2

Gauss’ law• Outside the hollow sphere with r a.

1. Draw a spherical surfaceof radius r around the charges.2. What is the charge enclosed?3. What is the total electric

E-r Graph

0 b a rba

E2 = 0

Electric potential

• Every test charge in an electric field possesses electric potential energy.

• The best way to describe the distribution of energy is using electric potential V.

• If the electric potential at a point in an electric field is V and we place a test charge q at the point, then the electric energy is

U = q.V

Electric potential• Work done on a test charge +q in a uniform

electric field E.

• Move a test charge +q from A to B against the electric force. We need to apply an external force F and the work done = F.d

1.External force = F2. Electric field strength E

3. DisplacementAB = d.

Electric potential• Find the least work done W.

1.External force = F2. Electric field strength E

3. DisplacementAB = d.

W = qEd

• The charge +q has gained an electric potentialenergy U = qEd.

Electric potential• The charge +q has gained an electric potentialenergy U = qEd.• What would happen to the charge if it is released

1. Electric field strength E

2. Charge +q

Electric potential• Electric potential V at a point is the work done per

coulomb required to bring a positive charge from infinity to the point.

• The electric potential at infinity is zero.

2. A varying force F to bring thecharge from infinity to this point X.

1. A +q charge at ∞

3. Work done on the charge = W4. The electric potential energy of the charge U = W.5. Electric potential at X is

Electric potential V

• In practice, the electric potential at a point is often referred to earth rather than infinity.

• In practice, the electric potential on the ground is zero.

Electric potential V

• The unit of V is J C-1 (joule per coulomb) or V (volt).

• The electric potential energy U of a charge q at a point of electric potential V is

U = q.V

Electric potential difference ΔV

• ΔVBA = VB - VA

Electric potential energy difference ΔU

• ΔUBA = UB - UA

Electric potential difference ΔV

• ΔVBA = VB – VA

• The potential between two points is the work done per coulomb on a positive charge in moving between them.

Example 8• Note that the direction of motion.

ΔVAB = - ΔVBA • If ΔU > 0, there is a gain in electric

potential energy. If ΔU < 0, there is a loss in electric potential

energy.• What would happen if the proton is allowed

to move freely?

Potential difference of parallel plates

• The potential difference V, the electric field E and the separation of the parallel plates are closed related.

1. Set the potentialto be zero.

2. Potential = V

3. Electric field strength = E

d4. Separation = d

Potential difference of parallel plates

• Find the potential difference between plates.

1. Place a charge +qon the side with 0V.

2. Move the charge to the other plate. What is the minimum force F needed?

3. Find the work doneand calculate V.

Potential between parallel plates

• Find the potential difference between plates.

1. Place a charge +qon the side with 0V.

2. Move the charge to a point with distance r from the plate. What is the minimum force F needed?

3. Find the work doneand calculate V.

drrEV ..

Potential between parallel plates

• Find the potential at any point between parallel plates.

VrrEVr ..

2. V changeslinearly betweenthe parallel plates

electronvolt (eV)

• 1 eV = work done on an electron in moving it through a potential difference of 1 V.

• It is a unit of energy.

• It is commonly used to measure the energy of particle

Example 9

• Ignore the gravitational force. It is too small

in comparison with the electric force.

electronvolt (eV)• Find the absolute value of 1 eV in terms of t

he unit J.

1. What is the charge q of an electron?

2. Move it through apotential difference of 1 V.

ΔV = 1V 3. Calculate U from U = qV.

4. 1 eV = 1.6 × 10-19 J

Potential due to a point charge

2. Take a charge q atinfinity.

3. Bring the charge q to this position at a distance r from the charge Q.

4. Calculate the workdone W and V.

1. There is a pointcharge Q.

Potential due to a point charge

Example 10

• 1 kV = 1000 V

• The potential may either be positive or negative depending on the point charge Q.

Potential due to a charged metal sphere

• The charge Q of the sphere resides on the surface.

1. Outside the sphere, the chargedsphere is similar to a point charge. What is the potential at this point?

2. What is thepotential on the surface of the sphere? R

3. Constant potential insidethe metal sphere!

Potential due to a charged metal sphere

Example 11

• Note that the potential of a metal sphere is constant inside in electrostatics.

• Two cases, outside or inside the sphere, must be considered.

Electric field strength and potential

• Field strength = rate of change of potential• The negative sign indicates the direction of electric field strength• Electric field strength is also called potentialgradient.

Electric field strength and potential

• Find the electric field strength due to a point charge

• Find the electric field strength due to a charged metal sphere

Inside the sphere Outside the sphere

Equipotential surfaces

• An equipotential surface is an imaginary surface on which the potential is the same.

• Work done is zero to move a test charge on the equipotential surface.

• Electric field lines are always normal to the equipotential surfaces.

Equipotential surfaces

Plotting equipotential lines

Example 12

Note that E = 0 if V is constante.g. inside a metal sphere.

Influence on the potentialof a charged object

• The potential of a charged object is affected by objects nearby.

• We study an example of a positively charged conducting sphere A.

• By an earthed conductor

RA - - -

2. By induction, thereare negative charges

3. The potential ofthe earthed conductoris zero

1. The potentialwithout the earthedconductor

4. The negative chargesreduces the potential.The potential with theearthed conductor. 0

• By an uncharged conductor

2. By induction, thereare negative and positive charges

1. The potentialwithout the uncharged conductor

4. The potential of theconductor is constant.

3. The potential withthe uncharged conductor

• By an uncharged conductor

3. The positive induced charges raises the potential..

1. The potentialwithout the uncharged conductor

2. The negative chargesreduces the potential.

• By a positively charged conductor

2. The positive charges are repelled to the far side.

1. The potentialwithout the charged conductor +

4. The potential of theconductor is constant.3. The potential with

the uncharged conductor.The potential is raisedby the positive chargeson the conductor.

• By a negatively charged conductor

2. The negative charges are attracted to the near side.

1. The potentialwithout the charged conductor

4. The potential of theconductoris constant and negative.3. The potential with

the uncharged conductor.The potential is reducedby the negative chargeson the conductor. 0

• By a flame probe (an isolated conductor in flame)

1. The potentialwith and withoutthe flame probe.

2. The flameprobe is neutraland does notaffect the potential

Influence on the potential of charged parallel plates

+Vo0 V

1. A pair of parallelplates at constant potential differenceand separation d.

2. The change of potentialinside is linear.

Influence on the potential of charged parallel plates

+Vo0 V

1. Place a metal blockof thickness b between the plates

2. The potential of themetal block is a constant.

3. If the block is exactlyat the middle, its potentialis

Flame probe

Flame probe1. By induction, there are induced charges (including negativeand positive charges) on the isolated conductor. 2. The flame produces ions (including positive and negativeions) to neutralize the induced charges.

2. Ions produced bythe flame

1. Induced charges

3. It is neutral

Flame probe investigation• As a flame probe does not affect the

potential, we can use it to measure the potential at a point.

• The potential at the tip of the needle is shown from the angle of deflection of the gold leaf. The higher the potential, the bigger the angle of deflection.

θmetalrod

gold-leaf

Flame probe investigation

• Move the flame probe away from the conducting sphere.

1 outside a charged metal sphere

connected to gold-leaf electroscope Metal sphere

at high potential

• What would be the result if the probe is not in flame ?

(Hint: The probe is an uncharged conductor

if it is not in flame.)

• Move the flame probe between two parallel plates.

• rV

connected to gold-leaf electroscope

Ice-pail experiment

1. A positively charged metal sphere A.

2. A neutral metal container B with a narrow mouth

3. A gold-leaf electroscopeto measure the charges

Ice-pail experiment

1. A positively charged metal sphere A.

2. A neutral metal container B with a narrow mouth

3. A gold-leaf electroscopeto measure the charges

4. The container Band the electroscopeare neutral. The gold-leafdoes not deflect.

Ice-pail experiment

1. Lower the positively charged metal sphere A into the containerB without touching it.

2. Induced negative chargesinside and equal amount of induced positive charges outside.3. A gold-leaf deflects

because of the inducedcharges.

Ice-pail experiment

1. Let A touch the inside of B.

2. The metal sphere losesall its charges. The inducedcharges are also neutralized.

3. The deflection of gold-leaf does not change because the induced positivecharges do not change.

Ice-pail experiment

1. Move A away. It is neutral.

2. All charges of A hasbeen transferred to Band the electroscope.

3. The deflection of gold-leaf does not change because the induced positivecharges do not change.

Ice-pail experiment

We may conclude that when we put a charge to touch the insideof a metal container, the charge will distribute completely tothe outside surface of the metal container.

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