cahier physique - physics workbook

in construction, please be patient.

source: more on potentials

PART I: WORK IN PHYSICS -

1) In 10th grade you learnt that the work done on a system is the force time the displacement. W = F x d
This equation works if : a) the force is constant b) the force is // to the displacement.
Example: A rock falling a height h will have an energy of weight x h = m g h
Example A hokey puck pushed on ice with a force F over a distance d will have an energy = F x d
For work to be done on a system, there must be a change in energy (for the system).
The work computes how efficient a force is in changing the energy of a system.
If the force is perpendicular to the displacement, no work is done. (you carry a suitcase, you walk at
a constant speed, the suitcase stays at the same level, you are not doing work on it)

So far we only considered constant force parallel to displacement. A constant force can make an angle with the displacement
so we need another formula The work done from A to B by a force F is W_AB = F AB cos(α) α is the angle between the vector F and AB.
In mathematics, F AB cos(α) is the dot product between the vector F and AB.

2) A point moves from A to B because of a constant force F. The force makes an angle of 30 degrees with the displacement AB.
A) Find the work done by the force. F = 100N and the magnitude of AB is 12cm.

B) Find the work if the angle is now 180 degrees. (F and AB have opposite direction). Why is the sign negative ? What does it mean ?

C) Find the work done if F and AB are prependicular.

3) The trajectory between A and B does not need to be a line. could be any kind of curve C along which the force is acting. The force F can also vary along the
path. In that case, the path has to be broken into tiny lines and the work done is the sum of the dot products:

The work becomes the line integral of the force along the path C. (sum of the dot product between F and the line element ds = integral)
Luckily the electric force acting on a charge q (qE) or the gravitational force acting on a mass m (mg) are conservative forces.
E is the electric field and g the gravitational field.
It means the work done by these forces only depends on the end points A and B and is independent of the path that connects A and B.
So if we need to compute the work done by a conservative force (line integral of E or g) between A and B ,
we can choose a convenient path that makes the line integral easy to evaluate.
The Energy of a mass or a charge in a conservative field force depends only on the location of the mass or the charge in the field.
So if A = B (closed curve) no energy can be extracted from the electric field or the gravity field.
The magnetic force acting on a moving charge is perpendicular to the displacement, so no work is done by the magnetic field.

The electric field E (Fe=qE) and the gravitational field g (Fg=mg) , because they are conservative, make the following statement true:
V= - sum (E.ds) between A and B = - line integral (E.ds) between A and B. V is the electric potential (energy per unit charge)
E.ds is the dot product between the vector E and the line element ds from the path AB. AB can be a segment since the sum (or line integral) is path independent.

Likewise : U = - Sum(g.ds) . U is the gravitational potential (energy per unit mass).
V and U are scalar field. (note: you can also write g =- grad U and E = - grad V see chapter vector analysis)

we will study 2 examples: POTENTIAL/ ENERGY in an uniform electric/gravitational field
and POTENTIAL in a field produced by a point charge/point mass.

PART2 : WORK DONE IN AN UNIFORM GRAVITATIONAL FIELD / UNIFORM ELECTRIC FIELD

Let's consider first a mass m in an uniform gravitational field. (see picture). Close from the ground, the field is constant.
The gravitational force Fg =m g and g = - 9.8m/s/s

Energy(B) - Energy(A) = - sum(Fg.ds) from A to B = - sum ( mg.ds) = - mg. (ds) = - mg.AB = - m |g| (h_A -h_B )
Energy(A) - Energy (B) = m |g| (h_A - h_B) and if we choose Energy(h=0) = 0 then Energy (P ) = m |g| h_P
or U(P) = |g|h_P U(P) is the gravitational potential at point P (energy per unit mass)
The gravitational potential energy a P in an uniform field depends only on the altitude of the point P.
(we knew that already) Any plane // to the ground is an equipotential surface. The energy of the mass does not change on that surface.
The field is perpendicular to these equipotential planes. This is always the case when the field force is conservative. The field is perpendicular
to the equipotential surfaces and shows the direction of the greatest rate of change of the energy.
----------------------------------------------------------------------------------------------------------------------------------------------
Let's compute the difference of potential in an uniform electric field E between 2 plates of a capacitor. we consider a positive unit charge.
VB - VA = - sum(E.ds) = - E D so VA - VB = ED or Vpositive- Vnegative = V_PN E D
Vplus - Vminus is the tension between the 2 plates or the difference in electric potential

Example : A dust charged dust q=10^-8C moves from the plate N to P. V_PN = 1000V.

Find the work done by the electric force. Why is the work negative ?
Note:

At the edges the field lines are no longer // . In blue the equipotential lines, always perpendicular to the field lines.

APPLICATION: WORK THEOREM

change in kinetic energy of an object of mass m = work done by outside forces .
0.5 m V_B² - 0.5 mV_A²= Sum W_AB (Fext)

1) An electron comes out of the Plate N and is accelerated to plate P by the field E . It is initially at rest. the mass is m = 9.1 10^-27 kg.
The voltage is U_PN = 2000V . The distance between the plates is d = 7cm. Find the maximu speed of the electron when it reaches
plate P.

PART 3: gravitational potential ( radial field) produced by a point mass M (or a planet M)

A mass M produces a gravitational field |g | = G M/r² where R is the distance from the mass. The vector field g is
is directed toward the mass M (black arrows) and its magntitude decreases with the distance squared. G is the gravitational constant.
Let's take the gravitational potential (energy per unit mass) to be zero very far from M (we say infinity, but it is exagerated).
so U(far) = 0. Let's consider the difference in potential between far and the point P located at a distance r from the mass M.
U(P) - U(far) = - sum (g.dr) from far to P along an arrow. (we can choose the path we want). The line element dr is outward so g.dr < 0
g.dr = - GM/r² dr so U(P) - U(far) = + line integral ( GM/r₂ dr ) between P and far = - GM / R
So U(P) = - G M / R or Energy(P) = - GmM/R at point P . U(P) is the gravitational potential.
The potential depends only on the distance from the mass M. The equipotential are spheres and the field lines are perpendicular to the
equipotential surfaces. The energy is negative because it is a binding energy. Energy has to be supplied to " free" the mass from the field.
We talk about a potential well. (practice problems)

In red the equipotential surfaces, in black the field lines.
The potential depends only on the distance from the mass M. The equipotential are spheres and the field lines are perpendicular to the
equipotential surfaces. The energy is negative because it is a binding energy. Energy has to be supplied to " free" the mass from the field.
We talk about a potential well.

We talk about a potential well :

Note: that it becomes easier to get out of the well as we move away from the EArth. The escape velocity is 11km/s = 7mph.
Kinetic energy has to be given to the mass to escape the well. THe energy fills the well. Note that a mass coming from far
and being trapped into the well will reach the Earth with the same speed.
source and more on gravitational potential
(practice problems)
-----------------------------------
part4: electric field produced by point charge Q (or by a charged sphere Q)

Let's see the case of a unit positive charge in the electric field of a positive mass Q. The force is now repulsive.
(see first image).

Let's suppose the electric potential V(far) = 0 and let's find the electric potential at a point P located at a distance R from the point charge.
V(P) - V(far) = - sum(E.dr) along a blue arrow. dr is outward so E.dr = |E| dr =Ke Q /r² dr (Ke is the coulomb's constant).
V(P) - V(far) = - line integral ( Ke Q /r² dr) from far to P = Ke Q / R
so V(P) = Ke Q / R this is the electric potential. if Q > 0 the potential is >0 too. Electric energy (P) = Ke / R q Q
The equipotential surfaces are spheres around Q. E is perpendicular to the surfaces.
Q>0 this is not a well. Q tries to push away the charge q. If Q < 0 then we have a well.

The electric field inside a conductive charged sphere is 0. A charge will not feel a force.
The potential is constant inside the sphere and decreases as 1/r outside the sphere.

part5 More electric fields ( conductor without symmetry , dipoles )

electric field inside a conductor hollow or solid
The electric field in any charged conductor, hollow or solid, is zero. (could be a sphere or any shape like a can).
This can be shown by GAussian's law (see vector analysis) . If you choose a close surface insise the conductor,
the flux of E out of the close surface is 0 because there is no charge inside. Inside the potential is constant.
The faraday cage is based on that principle. It shield from elecrtic field. So a radio would not get any signal. except for very high frequency ones.

distrubution of charge/ electric field on a non symmetric conductor/lightning and discharge

The charges will spread on the surface. The charges spread uniformly only on sphere. For other shape,
you get more charges where the surface is more curvy (the curve is higher) . The electric field lines are perpendicular to the surface.

--------------------------------------------------------
Here is the demonstration:
Imagine you connect 2 spheres, a small of radius A and a large of radius B by a conductive string.
They are place far away.

Because they are connected, they have the same potential. (otherwise, charge would flow until the potential is the same).
Because they are far away, the field lines don't interact with each other and VA = QA / (4pi eo RA) and VB = QB / (4pi eo RB)
and since VA=VB then QA/QB = RA/RB BUT:
let's consider now the density of charge dA = QA/SA = QA / (4pi RA²) and dB=QB/SB = QB/(4piRB²)
dA/dB = QA/QB x RB²/RA² = RB/RA plug in QA/QB
So the density of charge of the small charge is larger than the density of charge of the large sphere.
If the radius of B is 5 times the radius of A, the density of charge is 1/5. or the density of A is 5 times greater.
So RA 5 times smaller than RB means dA (density of charges) 5 times larger.
So the density of charge is greater on more curved surface like spikes. Than means more electric field lines.
That means the charges feel s ronger force on the more curved part of the conductor.
The lightning is always attracted to pointy objects lir building, tree to you.
If you walk toward a geant Van Graaf generator, a spark will form from the generator to your noise or head.
If you spread you arm, the spark will jump to your finger.
----------------------------------------------------------------------------------------------------------
WE can compute the electric field produced by a non symmetric conductor using Gauss's law. (see vector analysis).
E = d / eo with d density of charge. (same than a charged plate).

The lightning occurs between the bottom of a cloud (negative) and the ground (positive). The bottom of the cloud can be considered as
the surface of a conductor. The electrons feel the electric field . When the discharge occurs, electrons knock off
an electron from the oxygen and nitrogen molecules. The electrons knocked off can knock off other electrons
and you get a chain reaction down to the ground. It is called an avalanche. It is a doubling. 1 electron gives 2 the 2 gives
2 more so 4 then you get 8 ... Nitrogen and oxygen molecules are ionized and when their electrons come back light
is emitted because the electrons lose energy when they go back to a bound state. Each time electrons lose energy,
a photon is emitted. In that case the photons are visible.
Discharge ionized channels are called leaders. The negative charged leaders that descend first are called the stepped leader.
It can branch into a number of paths. When the leaders reach the ground it bounces back and you get a return stroke
that will follow the channel of ionized air molecules This is when the light show happens. As the electrons move, the air is heated
air pressure wave is formed and that causes the thunder. more on wikipedia

first the stepped leaders then the return stroke.

Le'ts do some cumputation. How strong E , the electric field produced by the bottom of the cloud, has to be
to get a discharge. If you have a discharge an electron will ionize an air molecule. So the electron has to cover
a distance of about 1 micrometer(10^-6m). The ionization of oxygen molecule is 12.5eV (1eV = 1.6 10^-19 J)
The ionization of nitrogen molecule is 15 eV. Let's make it about 10eV. THat's a difference of potential of 10V.
So the electric field needs to be strong enough to give an electron an energy of 10V/C.
10 = E x distance with distance = 10^-6m so E = 10⁷ N/C
Because the bottom of the cloud is not flat and the ground is not flat discharge occurs around 3 10⁶ V/m

Say the bottom of the cloud is 1km away for the ground and E = 310⁶N/C for a discharge. compute
the voltage between the ground the the cloud. (use V = E x , x is the distance in meters).
--------------------------------------------------
So you need an electric field of at lead 3 10⁶N/C to get a discharge. So if you have a capacitor with a voltage of 300V,
what is the distance between the plate for which a discharge will occur ? ( 300= E x ) solve for x
The distance is acutally 3mm because the surface is not flat.
-------------------------------------------------
Consider the Van Graaf generator

The voltage is V = Q /(4pi eo R)
E = Q / (4pi eo R²)
so V = E R
for a discharge E = 3 10⁶ N/C
So fill the table : (use meters)

R (in cm)	V (in KV)	Q (in micro coulombs)
5cm (0.05)
50cm

Elmo fire and hedinburg and napoleon

electric field lines and potential lines created by dipoles.

Below: 2 charges, the field lines and the equipotential lines. In the center the potential is 0 along the vertical line.. So it takes no work to bring
a charge from far to the center of the dipole along that verticale. The electric field is not zero. But the component of the electric field along this vertical is 0/

2 likes charge below. the electric field is zero between the charges

PLAY WITH THESE APPLETS to see what happens when you have more than 1 charge.

PART6: (more advanced) How to compute the electric field, given the potential V
(you need calculus)
- dv/dx = Ex and -dV/dy = Ey and - dV/dz=Ez
See vector analysis for explanation and proof (you need multivariable calculus)

or If the variable is r (cylindrical coordinates), Er = - dV/dr
V = Q / (4 pi eo r ) so if Er = -dV/dr = + Q ( 4 pi eo r²)
Since it is a vector, we cna multiply both sides by r rooth r
E = + Q ( 4 pi eo r²) r

Example: V = 10⁵ x 0 <x < 0.01 can describes the potential of a capacitor.
A -| |- B so VA = o and VB = 1,000 volts. The energy of plate B is 1000V higher than plate A.
We can use the relationship Ex = - dV/dx to find Ex
Ex = - 10⁵ x (constant, toward the negative x )
(actually equals to Q/s / eo )

http://en.wikipedia.org/wiki/St._Elmo's_fire