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Lecture 1

PHY-2049C Lecture Notes - Lecture 1: Fermi Energy, Superposition Principle, Stationary Point


Department
Physics
Course Code
PHY-2049C
Professor
H.Prosper
Lecture
1

Page:
of 4
PHY 2049C – Dr. Prosper/Y. Hori/W. Roberts
Week 1 – 01 (Tuesday Lecture)
Electric Charge, Force, and Field Pt. 1
• Electric Charge: Qualitative Facts
-There are two kinds of charge: positive (+) and negative (-)
-Charge is CONSERVED
-Lightning involves the flow of charge
-Charge can be transferred between objects
-Charge can be displaced
-Like charges repel, unlike charges attract
-Charge is quantized (it occurs in distinct units)
• Electric Charge: Polarization
-A metal is a lattice of positively charged ions immersed in a gas of electrons that
flows freely through the lattice.
-If a positively charged rod nears the metal, the freely floating gas of electrons nears
the positively charged rod. This creates a negative pole (near the positively charged
rod) and a positive pole (away from the positively charged rod) in the metal. This
process is known as POLARIZATION.
-An insulator is a lattice of electrically neutral atoms.
-When a charged rod nears an insulator, it polarizes the insulator (creates positive and
negative poles); HOWEVER, this is much less polarized than the metal.
• Electron Gas Pressure (“an aside” so I am not certain if this will be on the test)
-The pressure of the gas of free electrons in a metal is given by the approximate
formula:
P = (2/5) * ρ * Ef
Where ρ is the electron density (number of electrons per unit volume)
and Ef is the Fermi Energy: an energy characteristic of the material (would be given)
•Coulomb’s Law
-Coulomb’s Law states that: two stationary point charges q1 and q2 a distance r apart
exert a force on each other of magnitude:
o|F(1,2)| = k * q1 * q2 / r^2 = |F(2,1)|
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-Unit of charge is coulomb (C)
o1 C = 6.25 * 10^18 elementary charges
oTherefore, one elementary charge = e = 1.60 * 10^-19 C
oOne electron = -e; one proton = +e
-Follows superposition principle: the electric force on a point charge is the sum of the
Coulomb forces on it due to all other point charges.
Coulomb Force Problems
1. Electric force due to a uniformly charged rod
Find the electric force on a point charge Q1 at point P = (x, y) due to a thin uniformly
charged rod. Take the origin at the center of the rod. The total charge on the rod is Q2.”
Step 1: Define an element of charge
Imagine we are picking a point on the rod for this part. The charge on this rod is
evenly distributed! So, to find how much charge this point (dq) has: we need to find
the ratio of the length of this point (dl) to the length of the rod (L) and multiply it by
the total charge Q2. So..
dq = Q2*dl/L
Step 2: Write down the force of dq on Q1.
For this, we use Coulomb’s Law. r represents the distance between the dq and P
and k is a constant.
F2,1(vector) = [ (k * Q1 * dq) / (r^2) ] * r (unit vector)
Notice that we multiply by r hat (the unit vector of r) to make Coulomb’s Law
give the vector instead of just the magnitude. The r hat gives the vector its direction.
Now let’s substitute r in terms of x,y (the location of point P) and l (the position
of dq where –L/2 <= l <= L/2). r would be the hypotenuse of the right triangle with
legs of length x-l (horizontal) and y (vertical). Thus:
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r (vector) = (x-l) i + y j
[Note: i and j are unit vector representing the x axis direction and y axis direction,
respectfully. This is a notation you should know, though it is commonly written with
a ^ instead of dots.]
Using Pythagorean Theorem, we get:
|r| = [ (x-l)^2 + y^2 ]^0.5
Substituting into our vector equation we get:
F2,1 (vector) = [ (k * Q1 * dq) / ( (x-l)^2 + y^2) ] * r (unit vector)
Step 3: Write down the unit vector
The unit vector r hat is defined as: r(vector) / |r|
r (hat) = [ (x-l) / ( (x-l)^2 + y^2)^0.5 ]*i + [ y / ( (x-l)^2 + y^2)^0.5]*j
Step 4: Write down the net electric force on the point charge at P
Substituting r hat into our electric force equation and rewriting we get:
F2,1(vector) = k * Q1 * Q2 ( I_x * i + I_y j)
I_x = (1/L) {integral from –L/2 to +L/2} (x-l)dl / [ (x-l)^2 + y^2 ]^3/2
I_y = (y/L) {integral from –L/2 to +L/2} dl / [ (x-l)^2 + y^2 ]^3/2
Take a moment here to see what we did. The integral sign is here in order to sum up
the charge of all the points along the charged rod. The bounds for the integral was
where l was defined along: -L/2 to L/2.
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