Download Lec Electrostatics.notebook

Lec Electrostatics.notebook
January 24, 2013
What happens here?
Electric Field Lines
3­D view of Electric Fields
Lec Electrostatics.notebook
January 24, 2013
Electric Field Lines
Electric Field Lines
Lec Electrostatics.notebook
January 24, 2013
Electric Field Lines
What is the force between two charges?
F = kq1q2/r2 where k = 9E9 =1/4πεo
εo = 8.85E­12
(permittivity of free space)
E = F/q (calculate Electric Field at
position r from a point charge)
E = kq1q2/r2 / q2
E = kq1/r2 r^
Lec Electrostatics.notebook
January 24, 2013
Hollow Sphere
What is the force between two charges?
F = kq1q2/r2 where k = 9E9 =1/4πεo
εo = 8.85E­12
(permittivity of free space)
E = F/q (calculate Electric Field at
position r from a point charge)
E = kq1q2/r2 / q2
E = kq1/r2 r^
Lec Electrostatics.notebook
January 24, 2013
1m
P
1µC
What is the E field at point P?
What is the force on a 2C charge at point P?
1m
1µC
1m
1µC
P
P
What is the force on a 2C charge at point P?
E = kq1/r2 r^
^
E = k(1µC)/(1m)2 r (do for each charge)
E = 9000 N/C (in the +x & +y direction)
E = √90002 + 90002
E = 12728 N/C (at 45o)
What is the E field at point P?
1µC
1m
Lec Electrostatics.notebook
January 24, 2013
1m
P
P
1µC
What is the E field at point P?
What is the force on a 2C charge at point P?
Since E = 12728 N/C (at 45o)
and E = F/q
1m
Then F = Eq = 12728 N/C * 2C
1µC
F = 25456 N
(could also do this problem rotated 45o CCW)
Charge density:
λ = Q/l (linear)
σ = Q/A (area)
ρ = Q/V (volume)
calculus: dq = λ dl dq = σ dA
dq = ρ dV
Lec Electrostatics.notebook
January 24, 2013
Given a charged hoop, what is the E field at point P?
P
Given a charged hoop, what is the E field at point P?
Think symmetry.
P
The components parallel to the hoop's plane cancel out, so only use the components perpendicular.
Lec Electrostatics.notebook
January 24, 2013
Given a charged hoop, what is the E field at point P?
r
Start with: E = kq/r2
R
dE = k dq / r2
θ
P
r
But we want the component perpendicular to the plane of the hoop.
dE = k dq / r2 cosθ
Given a charged hoop, what is the E field at point P?
r
dE = k dq / r2 cosθ
R
θ
cosθ = x/r
x
P
r
dE = k dq / r2 (x/r)
We don't like r, so use:
r = √x2 + R2
Lec Electrostatics.notebook
January 24, 2013
Given a charged hoop, what is the E field at point P?
r
dE = (k dq / r2) cosθ
R
θ
cosθ = x/r
x
dE = (k dq / r2) (x/r)
P
r
We don't like r, so use:
r = √x2 + R2
dE = (k dq / r2) (x/r) = kx dq / r3
If there's a uniform distribution, we could just integrate now and get Q for our charge.
Here's what we do to take into account charge density λ.
dq = λ dl
dE = kx λ dl / r3
dE = kx λ dl / (x2 + R2)3/2
Integrate both sides... dl = 2πR
and λ = Q/2πR
(note if x >>R)
E = kx Q / (x2 + R2)3/2
2002E1. A rod of uniform linear charge density λ = +1.5 x 10­5 C/m is bent into an arc of radius R = 0.10 m. The arc is placed with its center at the origin of the axes shown above.
a. Determine the total charge on the rod.
b. Determine the magnitude and direction of the electric field at the center O of the arc.
A proton is now placed at point O and held in place. Ignore the effects of gravity in the rest of this problem. d. Determine the magnitude and direction of the force that must be applied in order to keep the proton at rest. e. The proton is now released. Describe in words its motion for a long time after its release.
Lec Electrostatics.notebook
January 24, 2013
2002E1. A rod of uniform linear charge density λ = +1.5 x 10­5 C/m is bent into an arc of radius R = 0.10 m. The arc is placed with its center at the origin of the axes shown above.
a. Determine the total charge on the rod.
q = λl
l = 1/3 (2πr)
l = 1/3 (2π*0.10) = 0.21m
q = 1.5x10­5 C/m * 0.21m
q = 3.1x10­6 C
2002E1. A rod of uniform linear charge density λ = +1.5 x 10­5 C/m is bent into an arc of radius R = 0.10 m. The arc is placed with its center at the origin of the axes shown above.
b. Determine the magnitude and direction of the electric field at the center O of the arc.
dE = k dq /r2
The 'y' components cancel
The 'x' components don't
dEx = (k λ dl /r2) cosθ
dl = r dθ
dEx = (k λ r dθ /r2) cosθ
dEx = k λ cosθ dθ / r Integrate both sides
o
240
Ex = ∫k λ cosθ dθ / r o
120
240o
Ex = k λ sinθ / r |
o
120
E = 2.3x106 N/C
Lec Electrostatics.notebook
January 24, 2013
2002E1. A rod of uniform linear charge density λ = +1.5 x 10­5 C/m is bent into an arc of radius R = 0.10 m. The arc is placed with its center at the origin of the axes shown above.
A proton is now placed at point O and held in place. Ignore the effects of gravity in the rest of this problem. d. Determine the magnitude and direction of the force that must be applied in order to keep the proton at rest. F = qE
F = 1.6x10­19 * 2.3x106 N/C
F = 3.7x10­13 N
2002E1. A rod of uniform linear charge density λ = +1.5 x 10­5 C/m is bent into an arc of radius R = 0.10 m. The arc is placed with its center at the origin of the axes shown above.
A proton is now placed at point O and held in place. Ignore the effects of gravity in the rest of this problem. e. The proton is now released. Describe in words its motion for a long time after its release.
It will accelerate to the right, with a decreasing acceleration until a constant velocity is reached.
Lec Electrostatics.notebook
January 24, 2013
Here is a charged wire, what is the E field at point P?
What is the force on a 2C charge at point P?
P
Here is a charged wire, what is the E field at point P?
What is the force on a 2C charge at point P?
Charge density: λ = Q/l ⇒ Q = λl
calculus: dq = λΔl = λdl P
So given a horizontal bar of length=l, with a charge Q, what is the E field at position P?
Lec Electrostatics.notebook
January 24, 2013
We need a new approach. For this, we're going to use Gauss' Law.
P
l
Think about the E field. There's only so much to go around. As you get further away, it gets weaker but covers more area.
Lec Electrostatics.notebook
January 24, 2013
Think about the E field. There's only so much to go around. As you get further away, it gets weaker but covers more area.
The shape doesn't matter.
Think about the E field. There's only so much to go around. As you get further away, it gets weaker but covers more area.
The E field, times the area, is a constant.
This constant is called the Flux.
Φ = E * A
Lec Electrostatics.notebook
January 24, 2013
Φ = E * A
Select a sphere around the charge, and calculate Flux.
Φ = E * dA
dA = 4πr2
For a point charge,
E = kq/r2 = (1/4πεo)q/r2
Φ = ((1/4πεo)q/r2) * (4πr2) = q/εo
The net flux due to a number of charges contained within any closed area is:
Φ = 1/εo * Σ q
Φ = E dA
Φ = 1/εo * Σ q
E dA = 1/εo * Σ q
Gauss's Law
The trick here: the dA is the area of a symmetrical surface around the charge.
Called the Gaussian Surface.
Lec Electrostatics.notebook
January 24, 2013
Think about the E field. There's only so much to go around. As you get further away, it gets weaker but covers more area.
E is a vector, so we really want where E is perpendicular the surface. We could use trig, or just pick symmetrical surfaces.
For a point charge, the Gaussian surface is obvious.
Lec Electrostatics.notebook
January 24, 2013
For a plate, the Gaussian surface is easy, once you know the secret.
For a plate, the Gaussian surface is easy, once you know the secret.
Having one surface within a conductor is convenient, too.
You want shapes that have sides parallel to the electric field lines, or perpendicular to them.
Lec Electrostatics.notebook
January 24, 2013
How much charge is contained?
­Q
­Q
­Q
­Q
What is the E field at point P, 2m from the long, charged wire?
P
λ = 0.10 C/m
Lec Electrostatics.notebook
January 24, 2013
What is the E field at point P, 0.4 m from the large, charged plate? It has a uniform electric field upward.
P
σ = 0.001 C/m2
This 0.1m radius, solid, non­conducting sphere has a uniform charge density ρ = 0.03 C/m3. What is the E field at the center of the sphere? What is the E field at r = 0.05m? What is the E field at the surface of the sphere? What is the E field at r = 0.2m?
r