8.022 (E&M) - Lecture 2
Topics:
Energy stored in a system of charges
Electric field: concept and problems
Gauss’s law and its applications
Feedback:
Thanks for the feedback!
Scared by Pset 0? Almost all of the math used in the course is in it…
Math review: too fast? Will review new concepts again before using them
Pace of lectures: too fast? We have a lot to cover but… please remind me!
Last time…
Coulomb’s law:
q1q 2
rˆ2 1
| r2 1 | 2
F2 =
Superposition principle:
i= N
FQ =
∑
FQ =
∫
i =1
V
qiQ
rˆi
| ri | 2
qdq
2
q3
8.022 – Lecture 1
qN
rQ
rˆi
qi
ρ dV Q
r̂
|r| 2
September 8, 2004
q1
V
rˆi
q5
Q
2
1
Energy associated with FCoulomb
y
q
Q
r2
x
r1
How much work do I have to do to move q from r1 to r2?
September 8, 2004
8.022 – Lecture 1
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Work done to move charges
How much work do I have to do to move q from r1 to r2?
W = ∫ FI • ds where FI = − FCoulomb = −
Assuming radial path:
W ( r1 → r2 ) = ∫ FI • ds = − ∫
r2
r1
Qqrˆ
.
r2
ds
q
Qqrˆ
Qq Qq
−
idrˆ =
2
r
r2
r1
Q
Does this result depend on the path chosen?
r2
r1
No! You can decompose any path in segments // to the radial direction
and segments |_ to it. Since the component on the |_ is null the result
does not change.
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8.022 – Lecture 1
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2
Corollaries
The work performed to move a charge between P1 and P2 is the
same independently of the path chosen
W12 = ∫
F • ds
=∫
F • ds
=∫
F • ds
Path1
Path 2
Path 3
y
P2
1
2
3
P1
x
The work to move a charge on a close path is zero:
W11 =
∫
F • ds = 0
Any
In other words: the electrostatic force is conservative!
This8, will
the–concept
September
2004allow us to introduce
8.022
Lecture 1of potential (next week)
5
Energy of a system of charges
How much work does it take to assemble a certain configuration of charges?
y
q2
q3
q1
x
P1
W (Q) = ∫ F • ds = 0 no other charges: F=0
∞
q1q2
r12
qq qq q q
= W1+ 2 + W1+3 + W2+3 = 1 2 + 1 3 + 2 3
r12
r13
r23
W1+ 2 = ∫ FI • ds =
W1+ 2+3
September 8, 2004
8.022 – Lecture 1
Energy stored by N charges:
U=
1 i=N
∑
2 i =1
j=N
qi q j
j =1
j ≠i
rij
∑
6
3
The electric field
Q: what is the best way of describing the effect of charges?
FES
1 charge in the Universe
2 charges in the Universe
Fq =
q
Q
qQ
rˆ
| r |2
But: the force F depends on the test charge q…
define a quantity that describes the effect of the charge
Q on the surroundings: Electric Field
E =
September 8, 2004
Fq
q
=
Q
rˆ
| r |2
Units: dynes/e.s.u
7
8.022 – Lecture 1
Electric field lines
Visualize the direction and strength of the Electric Field:
Direction: // to E, pointing towards – and away from +
Magnitude: the denser the lines, the stronger the field.
Demo
Sink
Faucet
+
-
Properties:
Field lines never cross (if so, that’s where E=0)
They are orthogonal to equipotential surfaces (will see this later).
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8.022 – Lecture 1
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4
Electric field of a ring of charge
Problem: Calculate the electric field created by a uniformly
charged ring on its axis
Special case: center of the ring
General case: any point P on the axis
P
z
R
Answers:
• Center of the ring: E=0 by symmetry
Qz
• General case:
E =
zˆ
3
(R 2 + z2 )2
September 8, 2004
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8.022 – Lecture 1
Electric field of disk of charge
Problem:
Find the electric field created by a
disk of charges on the axis of the disk
Trick:
a disk is the sum of an infinite number
of infinitely thin concentric rings.
And we know Ering…
P
z
R
r
(creative recycling is fair game in physics)
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8.022 – Lecture 1
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5
E of disk of charge (cont.)
Electric field of a ring of radius r:
zQ
E rin g ( r ) =
(r 2 + z 2 )
3
2
P
zˆ
z
If charge is uniformly spread:
d q = σ d a = 2π rσ d r
R
Electric field created by the ring is:
dE =
zσ 2 π r d r
3
zˆ
(r 2 + z 2 ) 2
Integrating on r: 0 R:
r=R
E =
∫
r=R
∫
dE =
r =0
r =0
r
⎛ 1
zˆ = 2πσ zzˆ ⎜
−
3
|z|
2
2 2
⎝
(r + z )
zσ 2π dr
September 8, 2004
⎞
⎟
2
2
R +z ⎠
1
8.022 – Lecture 1
11
Special case 1: R infinity
⎛ 1
−
For finite R: E = 2πσ zzˆ ⎜
⎝|z|
⎞
⎟
R +z ⎠
1
2
2
P
What if R infinity? E.g. what if R>>z?
Since
lim
R→∞
1
R2 + z2
=0
z
R
E = 2πσ zˆ
Conclusion:
Electric Field created by an infinite conductive plane:
Direction: perpendicular to the plane (+/- z)
Magnitude: 2πσ (constant!)
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6
Special case 2: h>>R
⎛ 1
−
For finite R: E = 2πσ zzˆ ⎜
⎝|z|
P
⎞
⎟
2
2
R +z ⎠
1
z
What happens when h>>R?
R
Physicist’s approach:
The disk will look like a point charge with Q=σπr2
E=Q/z2
Mathematician's approach:
Calculate from the previous result for z>>R (Taylor expansion):
2 −1 / 2 ⎞
⎛
⎞
1 ⎜⎛
⎛R⎞ ⎞
⎟
1 − ⎜1 + ⎜ ⎟ ⎟
⎟ = 2πσ zzˆ
⎜
⎟
⎟
z⎜
z
⎝
⎠
R2 + z2 ⎠
⎝
⎠
⎝
⎠
2
2
⎛
Q
1⎛R⎞ ⎞
⎛R⎞
~ 2 πσ zˆ ⎜ 1 − (1 − ⎜ ⎟ ) ⎟ = πσ zˆ ⎜ ⎟ = 2 zˆ
⎜
⎟
2 ⎝ z ⎠ ⎠ 8.022 –⎝Lecture
z ⎠ 1 z
September ⎝8, 2004
⎛ 1
E = 2 πσ zzˆ ⎜
−
⎝|z|
1
13
The concept of flux
v
n̂
Consider the flow of water in a river
The water velocity is described by
v ( x, y, z ) ≡ vx xˆ + v y yˆ + vz zˆ ≡ (vx , v y , vz )
Immerse a squared wire loop of area A in the water (surface S)
Define the loop area vector as A ≡ Anˆ
Q: how much water will flow through the loop? E.g.:
What is the “flux of the velocity” through the surface S?
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7
What is the flux of the velocity?
It depends on how the loop is oriented w.r.t. the water…
Assuming constant velocity and plane loop:
. 1 ) if A ⊥ v → Φ v = 0;
. 2 ) if A v → Φ v = vA ;
. 3 ) if A v = θ → Φ v = vA co s θ = v i A .
n̂
v
n̂
v
n̂
v
θ
General case (definition of flux):
Φv =
September 8, 2004
∫
S
v idA
8.022 – Lecture 1
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F.A.Q.:
what is the direction of dA?
Defined unambiguously only for a 3d surface:
At any point in space, dA is perpendicular to the surface
It points towards the “outside” of the surface
Examples:
.
.
.
n̂
Intuitively:
“dA is oriented in such a way that if we have a hose inside the surface
the flux through the surface will be positive”
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8
Flux of Electric Field
Definition:
ΦE ≡ Φ =
∫
S
E idA
Example: uniform electric field + flat surface
Calculate the flux:
E
Φ = ∫ E idA = E i A = EA cos θ
θ
S
Interpretation:
Represent E using field lines:
ΦE is proportional to Nfield lines that go through the loop
NB: this interpretation is valid for any electric field and/or surface!
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ΦE through closed (3d) surface
Consider the total flux of E through a cylinder:
Φ tot = Φ1 + Φ 2 + Φ 3
E
1
Calculate Φ1, Φ2, Φ3
dA
dA
2
3
Cylinder axis is // to field lines
Φ2=0 because E ⊥ nˆ
| Φ1|=|Φ3| but opposite sign since
Φ = ∫ E idA = EA cos θ
S
The total flux through the cylinder is zero!
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9
ΦE through closed empty surface
Q1: Is this a coincidence due to shape/orientation of the cylinder?
Clue:
Think about interpretation of ΦE: proportional # of field lines
through the surface…
Answer:
No: all field lines that get into the surface have to come out!
Conclusion:
The electric flux through a closed surface that does not contain
charges is zero.
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ΦE through surface containing Q
Q1: What if the surface contains charges?
Clue:
Think about interpretation of ΦE: the lines will either originate in the
surface (positive flux) or terminate inside the surface (negative flux)
Sink
Faucet
+
-
Conclusion:
The electric flux through a closed surface that does contain a net
charge is non zero.
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10
Simple example:
ΦE of charge at center of sphere
+Q
Problem:
Calculate ΦE for point charge +Q at the center of a sphere of radius R
Solution:
.E // d A e v e ryw h e re o n th e sp h e re
.P o in t c h a rg e a t d is ta n c e R : E = Q2
R
Φ =
∫
S
E idA =
∫
S
Q
Q
dA = 2
2
R
R
September 8, 2004
∫
S
dA =
rˆ
Q
4π R 2 = 4π Q
R2
8.022 – Lecture 1
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ΦE through a generic surface
What if the surface is not spherical S?
Impossible integral?
Use intuition and interpretation of flux!
Version 1:
S
+Q
Consider the sphere S1
Field lines are always continuous
S1
ΦS1=ΦS=4πQ
Version 2:
Purcell 1.10 or next lecture
Conclusion:
The electric flux Φ through any closed surface S containing a net charge Q
is proportional to the charge enclosed:
Φ=
September 8, 2004
∫
S
E i dA = 4π Qenc
8.022 – Lecture 1
Gauss’s law
22
11
Thoughts on Gauss’s law
Φ=
∫
S
E idA = 4π Qencl (Gauss's law in integral form)
Why is Gauss’s law so important?
Because it relates the electric field E with its sources Q
Given Q distribution
find E (integral form)
Given E
find Q (differential form, next week)
Is Gauss’s law always true?
Yes, no matter what E or what S, the flux is always = 4πQ
Is Gauss’s law always useful?
No, it’s useful only when the problem has symmetries
September 8, 2004
8.022 – Lecture 1
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Applications of Gauss’s law:
Electric field of spherical distribution of charges
Problem: Calculate the electric field (everywhere in space) due to a
spherical distribution of positive charges or radius R.
y
(NB: solid sphere with volume charge density ρ)
Approach #1 (mathematician)
• I know the E due to a point charge dq: dE=dq/r2
• I know how to integrate
• Solve the integral inside and outside the
sphere (e.g. r<R and r>R)
r '= r
∫
r '= 0
r '= r
dE =
dq
=
r'
r '= 0
∫
r '= r
∫
r '= 0
ρ dV
r '2
= ∫ dθ ∫ dφ
r '= r
∫
r '= 0
ρ r '2
r ''2
+
R
x
sinθ d θ d φ
Comment: correct but usually heavy on math!
Approach #2 (physicist)
• Why would I ever solve an integral is somebody (Gauss) already did it for me?
• Just use Gauss’s theorem…
Comment: correct, much much less time consuming!
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Applications of Gauss’s law:
Electric field of spherical distribution of charges
Physicist’s solution:
1) Outside the sphere (r>R)
Apply Gauss on a sphere S1 of radius r:
Φ =
∫
S1
S2
r
E i dA = 4π Q enclosed
+
S ym m etry: E is constant on S1 and
2
∫ E i dA = E 4π r = 4π Q
to dA .
R
→
E =
For r>R, sphere looks
like a point charge!
Q
r2
S1
r
S1
2) Inside the sphere (r<R)
Apply Gauss on a sphere S2 of radius r:
A g ain : Φ =
∫
S2
∫
S2
E i d A = 4 π Q en clo se d ; sym m e try: E is c o n s ta n t o n S 2 a n d
E i d A = E 4π r 2 ;
Q en c =
September 8, 2004
∫ ρ dV
=ρ
4
π r3 →
3
E =
to d A .
4
πρ r
3
8.022 – Lecture 1
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Do I get full credit for this solution?
Did I answer the question completely?
No! I was asked to determine the electric field.
The electric field is a vector
magnitude and direction
r
How to get the E direction?
+
Look at the symmetry of the problem:
Spherical symmetry
C om plete solution:
Q
E = 2 rˆ for r>R
r
4
E = πρ rrˆ for r<R
3
September 8, 2004
R
E must point radially
S1
r
E
R
8.022 – Lecture 1
r
26
13
Another application of Gauss’s law:
Electric field of spherical shell
Problem: Calculate the electric field (everywhere in space) due to a
positively charged spherical shell or radius R (surface charge density σ)
Physicist’s solution:apply Gauss
1) Outside the sphere (r>R)
Apply Gauss on a sphere S1 of radius r:
Φ =
∫
S1
+
E i dA = 4π Q enclosed
Sym m etry: E is constant on S1 and
∫
S1
→
S2
R
to dA.
S1
E i dA = E 4π r 2 = 4π Q encl = 4π σ ( 4π R 2 )
E =
4πσ R 2
Q
rˆ = 2 rˆ
r2
r
sam e as point charge!
1) Inside the sphere (r<R)
NB: spherical symmetry
E is radial
Apply Gauss on a sphere S2 of radius r. But sphere is hollow
September 8, 2004
Qenclosed=0
E=0
8.022 – Lecture 1
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Still another application of Gauss’s law:
Electric field of infinite sheet of charge
Problem: Calculate the electric field at a distance z from a positively
charged infinite plane of surface charge density σ
Again apply Gauss
Trick #1: choose the right Gaussian surface!
Look at the symmetry of the problem
Choose a cylinder of area A and height +/- z
Trick #2: apply Gauss’s theorem
Φtot = Φside + Φtop + Φbottom
Symmetry: E // z axis Φside=0 and Φtop = Φbottom
Φ =
∫
∫
cylinder
cylinder
z
E
E i dA = 4π Q enclosed
E i dA = 2 ∫ EdA = 2 EA = 4π (σ A )
September 8, 2004
top
8.022 – Lecture 1
→ E = 2πσ zˆ
28
14
Checklist for solving 8.022 problems
Read the problem (I am not joking!)
Look at the symmetries before choosing the best coordinate system
Look at the symmetries again and find out what cancels what and the
direction of the vectors involved
Look for a way to avoid all complicated integration
Remember physicists are lazy: complicated integral
somewhere or there is an easier way out!
you screwed up
Turn the math crank…
Write down the complete solution (magnitudes and directions for all the
different regions)
Box the solution: your graders will love you!
If you encounter expansions:
Find your expansion coefficient (x<<1) and “massage” the result until you
get something that looks like (1+x)N, (1-x)N, or ln(1+x) or ex
Don’t stop the expansion too early: Taylor expansions are more than limits…
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Summary and outlook
What have we learned so far:
Energy of a system of charges
Concept of electric field E
To describe the effect of charges independently from the test charge
Gauss’s theorem in integral form:
Φ. =
∫
S
E i dA = 4π Qencl
Useful to derive E from charge distribution with easy calculations
Next time:
Derive Gauss’s theorem in a more rigorous way
See Purcell 1.10 if you cannot wait…
Gauss’s law in differential form
… with some more intro to vector calculus… ☺
Useful to derive charge distribution given the electric fields
Energy associated with an electric field
September 8, 2004
8.022 – Lecture 1
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