Physics 11b
Lecture #2
Electric Field
Electric Flux
Gauss’s Law
What We Did Last Time
„
Electric charge = How object responds to electric force
Comes in positive and negative flavors
„ Conserved
„
„
Electric force
„
Coulomb’s Law
„
„
1 q1q2
Fe =
rˆ
2
4πε 0 r
Same inverse-square law as gravity
Sign makes the dynamics different
qi
rˆ
„ Superposition Principle FQ =
∑
2 iQ
4πε 0 i riQ
Q
Today’s Goals
„
Continue on the superposition principle
Work out a couple of problems
„ Electric dipole
„ Continuous charge distribution
„
„
Introduce electric field E
First (and the easiest) of the E&M fields
„ Field lines
„
„
Define electric flux ΦE
„
Gauss’s Law
Superposition Principle
„
Suppose we have many charges distributed in space
„
„
What is total force on charge Q
from all the other charges?
q
3q
Principle of Superposition
Just add them up
„ Force from each one not affected
by the others
-q
„
1 Qqi
FQ = ∑
rˆiQ
2
i 4πε 0 riQ
qi
=
rˆ
∑
2 iQ
4πε 0 i riQ
Q
2q
Q
-q
Force from the
i-th charge
-2q
q
Electric Dipole
„
Two charges +q and –q separated by d
„
„
q′
Bring another charge q’ nearby
„
„
Such pair is called an “electric dipole”
We call it a “test charge”
r
What’s the force on q’?
+q
„
−q
This is easy
„
Calculate the force from +q
„
What’s the distance between +q and q’?
Calculate the force from –q
„ Add the vectors
„
d
Electric Dipole
„
Force F1 from +q is
F1 =
„
„
1
4πε 0
(
qq′
r +( )
2
d 2
2
)
qq′
=
4πε 0 r 2 + d 2 4
1
2
q′
F
Same size for F2 from –q
Vector addition Î a horizontal F
„
F1
What’s the length?
F1 : F = r 2 + d 2 4 : d
qq′d
F=
4πε 0 (r 2 + d 2 4)3 2
1
r
F2
−q
+q
d
Continuous Distribution
„
Electric charge is often distributed over objects, e.g.
Along a length of wire (1-d)
„ On the surface of a metallic plate (2-d)
„ Inside the volume of a conducting ball (3-d)
„
„
Superposition principle needs to be modified
qi
FQ =
rˆ
∑
2 iQ
4πε 0 i riQ
Q
Q
dq
FQ =
rˆ
2
∫
4πε 0 r
Sum over individual charges
Integral over line/surface/volume
Uniform Linear Charge
„
Charge Q uniformly distributed on a thin rod of length ℓ
Test charge q is near the middle of the rod
y
„ Force on q?
„
q
r
x dx
x
„
Smart set up is everything
Define x-y coordinates
„ Pick a small piece of the rod and call it dx
„ Calculate the force from dx and integrate
„
A
Uniform Linear Charge
Q
dx
„ The piece dx has charge
A
„ Force from this piece is
1 QA dx ⋅ q
dF =
4πε 0 r 2 + x 2
„
Note that the force from the
green piece is same size
and flipped in x
„
Only the y-component will
remain after adding up
dFy =
dx ⋅ q
r
4πε 0 r + x 2 r 2 + x 2
1
Q
A
2
Linear charge density (C/m)
dF
q
r
x dx
A
y-component Î Integrate this!
Uniform Linear Charge
„
Integrate dFy
Qqr
dx
− A 2 4πε A ( r 2 + x 2 )3 2
0
F =∫
A2
1
1
Qqr A 2
=
dx
2
2 32
∫
2
−
A
4πε 0 A
(r + x )
y
dF
A little (?) math work here
Qq
1
F=
yˆ
2
2
4πε 0 r r + A 4
„
q′
r
x dx
A
This wasn’t easy – Is it really correct?
„
Check the direction, r-dependence, large-r limit
Electric Field
„
Look at the superposition principle again
What this charge is
qi
FQ =
rˆ
∑
2 iQ
4πε 0 i riQ
Q
What all the other
charges are and
where they are
Most of the “interesting” part has nothing to do with the
value of Q itself
„ But the position of Q is important – Let’s call it r
r − ri
ˆ
r
=
r
=
r
−
r
riQ = r − ri
iQ
iQ
i
r − ri
„
qi (r − ri )
FQ =
∑
4πε 0 i | r − ri |3
Q
A function of r
Electric Field
„
qi (r − ri )
How about writing FQ = Q ⋅ E(r ) = Q
∑
4πε 0 i | r − ri |3
1
„
E(r) is a vector function of position r
„
„
For each position in space, there is a vector E
Also called a “vector field”
It tells you “what will happen if you put a charge here”
„ We call it the electric field
„
„
„
Unit of E is Newton/Coulomb (N/C)
Simple case: only one charge q at r = 0
1 qr
1 q
E(r ) =
=
rˆ
3
2
4πε 0 r
4πε 0 r
Electric Field
„
E field created by a single charge
1 q
E(r ) =
rˆ E points outward
2
4πε 0 r
„
„
Guess why it’s called “field”
We’ve introduced a middle-man
for the electric force
Charge creates E field
„ E field creates force on charge
„ End result is unchanged (of course)
„
Field Lines
„
Tedious (and busy) to draw little arrows everywhere
„
„
Faraday came up with an easier
way = field lines
„
„
Took me 10 minutes to draw Î
Instead of little arrows, draw long,
continuous arrows originating
from the charge
Field lines represent direction
of the E field just as well as the
little arrows
„
What about the size of E?
Field Line Density
„
E field weakens with distance – Inverse-square law
1 q
E=
4πε 0 r 2
„
Field lines become less crowded
with distance
„
Consider a sphere at r
Surface area S = 4π r 2
„
Number of field lines N is constant
Density =
„
Nε 0
N
N
E
=
=
2
S 4π r
q
E
Density of field lines is proportional to the E field strength
Field Line Rules
„
You can often draw field lines without calculation
„
Just follow a few rules
¾ Field lines start from a positive charge and end in a
negative charge
Exception: it’s OK to come from/go to infinitely far away
¾ Field lines cannot split, merge, or cross each other
¾ The number of field lines attached to a charge is
proportional to the amount of the charge
„
The last rule ensures field line density ∝ E field
Examples with Two Charges
„
Examples from the textbook
Number of Field Lines
„
Imagine a small area of a surface in an electric field
Let’s call the area ∆A
„ It could be any shape, any angle
„
„
How many field lines run through this area?
„
∆A
E
Remember: density of
field lines ∝ E
„
„
Number of field lines
should be ∝ E × ∆A
Right?
Number of Field Lines
„
The answer depends on the angle between E and the
surface
„
Define a unit vector n standing normal to
the surface
∆A
θ
n
E
Projected in the direction of E, the area ∆A Î ∆A cosθ
„ Number of field lines ∝ E ∆A cosθ
„ We can also write this as E·∆A where ∆A ≡ ∆A n
„
„
We can integrate this to find the number of field lines
passing through any surface
Electric Flux
„
We now define electric flux as Φ E = ∫ E ⋅ dA
S
S is a surface area, dA = dA n
„ ΦE is proportional to the number of field lines going thru S
„
„
Unit of ΦE is Newton·m2/Coulomb (N·m2/C)
„
NB: sign of flux depends on the direction of n
„
That is, you must define which side of S is “positive”
S = 1 m2
n
ΦE = −1 Nm2/C
n
E =1 N C
ΦE = 1 Nm2/C
Simple Flux Example 1
„
Consider a sphere of radius r around a charge q
Define n to point outward
„ We know E and n are parallel
1 q
„ We also know E =
4πε 0 r 2
„
ΦE =
∫
E ⋅ dA = 4π r × E =
2
sphere
„
n
q
r
q
ε0
This is hardly surprising
ΦE should be proportional to the number of field lines
coming out of charge q, which should be proportional to q
„ We just didn’t know that the constant was 1/ε0
„
Simple Flux Example 2
„
Consider a sphere in a uniform E field
„ Take polar coordinates (θ, φ)
relative to the direction of E
Φ E = ∫∫ E cos θ r sin θ dθ dφ
2
r
n
θ
π
= 2π Er 2 ∫ cos θ sin θ dθ
0
=0
„
Again hardly surprising
Incoming E on one side is balanced by outgoing E
„ We know field lines never disappear without a charge
„
Gauss’s Law
„
Net flux through a closed surface is given by the net
charge inside the surface by
Φ E = v∫ E ⋅ dA =
qin
ε0
This seems natural, from what we’ve found so far
„ Formal proof is found in textbook 24.5
„
„
This is more useful than it looks
„
We will see in the next lecture
Summary
„
„
Introduced electric field E by FQ = Q ⋅ E(r )
Field lines and the rules
From a positive charge to a negative charge
„ No splitting, merging, or crossing
„ Number of field lines ∝ amount of charge
„ Density ∝ E field
„
„
Defined electric flux Φ E = ∫ E ⋅ dA
S
„
∝ number of field lines through the surface
„
Gauss’s Law
ΦE =
qin
ε0