Download Electrostatics

Electrostatics
Nicholas Kortendick
[email protected]
January 7th
TED Talk
Just how big are atoms?
If atoms are like blueberries in the earth, the nucleus would be invisible, because it is so small. If the
blueberry was the size of a football stadium, the nucleus would just be a marble. An electron is also very
dense.
Lecture
Protons and electrons carry charge, which is the same amount, despite the mass difference. A proton has
a charge of 1.6 × 10−19 coloumbs. An electron has the same charge, but negative. Any charge calculation
will be a multiple of 1.6 × 1019 .
A proton and an electron are attracted by an electrical forces. A protron pair and an electron pair
are repelled by an electrical force.
Lab
The second period was used to fill out the worksheet entitled Charge Exploration
January 8th
Lecture
Charge: The fundamental charge is 1.6 × 1019 Coloumbs. Total charge can be represented with
q = ne
Grounding: Provide a conducting path from charge objects to the ground Types of charging: Induction,
contact, and friction
Induction: In the presence of a charge object, neutral objects experience polarization. The polarized
conductor can then be charged by splitting into two pieces or grounding one side. No charge is transferred
from the initially charged object.
1
January 9th
Lecture
Why are you safe inside your care during a lightning storm?
It’s a faraday cage; the electrons travel around you
Coloumb’s Law is:
F E = k0
|q1 q2 |
r2
where
k0 = 9 × 109
Elementary Particle
Mass (kg)
Electron
9.109 × 10−31
Proton
1.673 × 10−27
1.675 × 10−27
Neutron
Charge (C)
−1.6 × 10−19
1.6 × 10−19
0
January 10th
Lecture
E=
X kQ
r2
An Electrical Field (N/C):
E=
FE
Q0
F = qE
Using point charges (spheres) and electrical fields, substitute to derive:
E=
kQ
r2
+Q E
2Q
The electrical field at the midpoint is two parts: E2
(0,0) x (L,0)
(left component) and E1 (right component). The net field is found to be:
Imagine the following situation
k2Q
+kQ
−
2
(L/2)
(L/2)2
−4kQ
L2
Imagine a situation with a particle 1 meter located up the y axis with charge of −3q and one 1 meter
up the x axis with charge -4Q. Calculate the electric field of a particle at the origin.
k3Q k4Q
+
1
1
2
January 11th
Lecture
Mapping Electric Field Rules:
1. Lines begin on positive charges (or infinity)
2. Lines end on negative charges (or infinity)
3. Lines Point in direction of force
4. Line density is proportional to field strength
5. Number of lines leaving or entering a charge is proportional to charge
See worksheet for graphing
January 14th
Lecture
If an electrical field is ”constant” you can use simple Kinematics. Additionally, if a problem does not
mention gravity, only focus on the electrical force. See the second web assign homework assignment
January 15th
Lecture
Electrical Fields Due to Continuous Charge distributions
Imagine an electron at the origin and a beam lying on the x origin starting at (a, 0) of length L. It has
a total charge of Q. We CANNOT treat the beam as a point particle. So to begin, pick an arbitrary
point, preferably not one on the end. The point has a thickness of dx. We know that the electrical field is
pointing to the left; how can we find its magnitude?
de =
kdq
x2
We also know that:
dx
dq
=
L
Q
So we should take the integral of the initial equation to form:
Z
Z E
kQ a+L dx
de =
L a
x2
0
kQ 1 a+L
[− ]a
L
x
kq a + L − a
E= [
]
L a(a + L
kq
E=
a(a + L)
E=
Now let’s have a point at (y, 0) and a beam beginning at the origin extending to (L, Q). This will be
diagonal. The process will be the same, but it will include some trigonometry.
dE =
3
kdq
r2
kdq
sinθ
r2
As we integrate, the angle changes and so does the vertical distance. We have to do some substitutions.
We know
3/2
y
−sinθ = 2
2
Y +x
−r2 = x2 + y 2
Q
−dq = dx
L
We substitute back in to form:
kQydx
dEy = k
L(y 2 + x2 )1/2
Z Ey
Z
Qy L
dx
dEy = k
L 0 (y 2 + x2 )3/2
0
dEy =
January 16th
Lecture
We will be solving with the following three relations:
λ= Q
σ=Q
ρ = VQ
L
A
We can prove the following:
EY Semi−Inf inite =
kλ
y
2kλ
y
Now we will solve a disk of charge through a secondary method. The radius is R, and it is centered a
distance Y away from a point. We will define the radius of a smaller, contained hoop to be r with a
kQy
thickness of dr. Additionally, E = (R2 +y
2 )1/2
EInf inite =
kdQy
(r2 + y 2 )3/2
dE =
dQ = σdA = σ2πrdr
Plugging in and integrating produces:
Z
dEy = kσπy
R
(r2 + y 2 )−3/2 2rdr
0
Then we can use the power rule to evaluate to:
1
1
Ey = 2kσπy( − p
)|R
0
y
R2 + y 2
For an infinite disk
E = 2kσπ
Very very far away it would start to behave like a point charge.
We will solve one final shape: a semicircle. Make sure to get the first line, part of the electrical field is
from a point.
kdq
dE = 2
R
4
dq = λds
ds = Rdφ
kλRdφ
cosφ
R2
Z
2kλ θ
cosφdφ
E=
R 0
dEx =
E=
2kλ
sinθ
R
5