Download Physics 227: Lecture 3 Electric Field Calculations

•
Physics 227: Lecture 3
Electric Field Calculations, Symmetry,
Charged Particle Motion, Flux, Gauss’s Law
Lecture 2 review:
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•
•
F1 on 2 = k q1 q2 /
•
Electric field determined by test charge: E = Fon qtest / qtest
•
Superposition: Forces are vectors, Coulomb forces add as
``Cheat’’ / formula sheet added
vectors.
to Sakai
resources - tell me about any typos
•
Electric fields are vectors, and add as vectors. Unique
direction and magnitude at each point in space.
•
Efrom point charge = k q / r2
Field lines show direction of electric field: E tangent to find
line, fields lines start on + charge, end on - charge, but may
start / stop at r = ∞, field strength ∝ number of field lines,
number of field lines ∝ charge.
Monday, September 12, 2011
Calculating the Electric Field
•
Superposition allows the electric field to be calculated for
an arbitrary fixed charge distribution ρ(x,y,z). Let:
� x)=
E(�
�
•
�
i
Or:
� x)=
E(�
�
qi
r̂i →
2
4π�0 ri
�
i
qi
r̂i →
2
4π�0 ri
�
ρ(�x)
ˆ
∆
d x
4π�0 |�x − �x� |2
�
1 ˆ
dq
∆
4π�0 r2
3
•
Where Δ is the unit vector from the position of the charge
to the position at which we are evaluating the electric field.
•
The integrals might require numerical evaluation, turning this
into a computation problem.
Monday, September 12, 2011
Electric Field of Two Charges
y
q1
q2
d
� x)=
E(�
�
�
i
•
•
•
qi
r̂i →
2
4π�0 ri
x
d
�
ρ(�x)
ˆ
∆
d x
4π�0 |�x − �x� |2
3
What is the electric field along the x axis?
For x > d:
For 0 < x < d:
Monday, September 12, 2011
q
q
2
1
�
x̂
+
x̂
E=
4π�0 |x − d|2
4π�0 |x + d|2
−q
q
1
2
�
E=
x̂ +
x̂
2
2
4π�0 |x − d|
4π�0 |x + d|
Notice that at
x >> d the
field falls like
the field of a
point charge
Electric Field of Two Charges
y
� x)=
E(�
�
E
�
i
-q1
q1
d
d
x
qi
r̂i →
2
4π�0 ri
�
ρ(�x)
ˆ
∆
d x
4π�0 |�x − �x� |2
3
•
What is the electric field along
the y axis?
•
You can see the vertical
components will cancel.
−2q
d
1
� =
�
E
x̂
×
4π�0 (y 2 + d2 )
y 2 + d2
−2q
d
1
� =
E
x̂
4π�0 (y 2 + d2 )3/2
Monday, September 12, 2011
Notice that at large y
the dipole field falls
faster than the field
of a point charge
Electric Field of Two Charges
• Why
y
does the field fall faster than the
field of a point charge?
E
A. It cannot - the professor is wrong!
-q1
q1
d
d
B. It always falls faster for 2 or more charges.
x
C. The two charges cancel, so the field is 0
everywhere!
D. Because the fields of the two charges
partially cancel.
E. It falls faster in some directions, but not all.
� =
E
Monday, September 12, 2011
Notice that at large y
−2q1 d
the dipole field falls
x̂
4π�0 (y 2 + d2 )3/2 faster than the field
of a point charge
Electric Field of Two Charges
• Why
y
does the field fall faster than the
field of a point charge?
E
A. It cannot - the professor is wrong!
-q1
q1
d
d
B. It always falls faster for 2 or more charges.
x
C. The two charges cancel, so the field is 0
everywhere!
A. Often, but not this time.
B. Not for, e.g., two same D. Because the fields of the two charges
partially cancel.
sign charges.
C. There is a partial, not E. It falls faster in some directions, but not all.
total, cancellation.
E. Dipole field falls like 1/r3,
Notice that at large y
−2q1 d
though you have not been �
the dipole field falls
E=
x̂
told enough so far to know
4π�0 (y 2 + d2 )3/2 faster than the field
of a point charge
this answer is wrong
Monday, September 12, 2011
Electric Field of a Ring of Charge
� x� ) =
E(�
�
i
qi
r̂i →
2
4π�0 ri
�
1 ˆ
dq
∆
4π�0 r2
Example 21.10
•
What is the electric field
along the x axis?
•
You can see the vertical
components will cancel.
•
The integrand does not
depend on position along
the ring, so the integral
can be done ``by
inspection’’
Notice that at large x >> a, this approaches the field of a point charge.
Monday, September 12, 2011
Symmetry Arguments
Sometimes you can determine something about the shape of
the electric field from symmetry arguments
For a point charge or a uniform sphere of charge, the field is
radial and spherically symmetric: E(r,θ,φ) → E(r)
+
+
+
The charge looks the same to both observers. The charge looks the same when
an observer rotates about its head. There is nothing to distinguish the observer’s
up from your down, or left from right. There is no preferred direction in space
except for the line going from the charge to the observer. The field must be
along this line. (You are a point in space, your body does not count!)
Monday, September 12, 2011
Symmetry Arguments
Sometimes you can determine something about the shape of
the electric field from symmetry arguments
For an infinitely long uniform line of charge, the field is radial
and perpendicular to the line: E(ρ,φ,z) → E(ρ)
+
+
+
+
+
+
+
+
+
+
+
+
φ
+
z
ρ
If you move in z, the line looks the same, so the field has no z component. If
you go around the line in the φ direction, the line looks the same, so the field
has no φ component. The field can only be in the ρ direction.
Monday, September 12, 2011
Symmetry Arguments
Sometimes you can determine something about the shape of
the electric field from symmetry arguments
For an infinite uniform plane of charge, the field is
perpendicular to the plane: E(x,y,z) → E(z)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
y
+
+
+
+
+
If you move in x or y, or if you rotate about the z axis,
the plane looks the same, so the field has no x or y
components. The field can only be in the z direction.
Monday, September 12, 2011
x
Motion in an Electric Field
•
Soon we will learn that the field inside a ``parallel plate
capacitor’’ is ≈ constant, as long as the size of the plates (√A) is
much greater than the separation of the plates (d)
x
•
+
v0
y
Since the E-field is
constant, the electric
force F = qE is constant
between the plates.
• Thus, F = ma leads to ay = qE/m and ax = 0.
• With v(t=0) = v0x, we find that x = v0xt, and y = (1/2)at2 = (1/2)(qE/m)t2.
• Put t = x/v0x into the equation for y to obtain: y = (1/2)(qE/mv0x2)x2.
• This is a parabolic trajectory, just like motion of a projectile in a
uniform gravitational field. It does not follow the field lines.
Monday, September 12, 2011
Introduction to Flux
•
For a constant E-field, flux through a surface is, crudely speaking,
the number of field lines through the surface.
•
Toy problem: inside the parallel plate capacitor the E field is constant
and vertical, 25 field lines per m2
• There is a 1 m2 horizontal surface in side the capacitor.
• What is the flux through this horizontal surface?
• Flux = field line density x area = 25 field lines/m2 x 1 m2
Monday, September 12, 2011
= 25 field lines
Toy Examples, with flux of water
Velocity of water
is a vector field,
like electric fields.
• Water
of density 1000 kg/m3 flows at 3 m/s through a horizontal
pipe of cross sectional area 1 m2. What is the flux of water through
a vertical plane through the pipe?
•
Flux = 1000 kg /m3 x 3 m/s x 1 m2 = 3000 kg/s.
3 flows at 3 m/s through a horizontal
Water
of
density
1000
kg/m
•
pipe of cross sectional area 1 m2. What is the flux of water through
a horizontal plane through the pipe?
•
Flux = 0. No water goes through the plane! Water flows along the
plane, but it does not cross over it.
Monday, September 12, 2011
More toy examples
• Water
of density 1000 kg/m3 flows at 3 m/s through a U-shaped
horizontal pipe of cross sectional area 1 m2. What is the flux of
water through a vertical plane through the upper+lower pipes?
•
•
•
Fluxupper = 1000 kg /m3 x 3 m/s x 1 m2 = 3000 kg/s.
Fluxlower = 1000 kg /m3 x -3 m/s x 1 m2 = -3000 kg/s.
Fluxtotal = Fluxupper + Fluxlower = 0
Monday, September 12, 2011
More formal definition of Flux
E-field lines
ΦE =
ˆ
Surface: A
unit vector
⊥ to surface
try
ΦE =
�
�
� · dA
�=
E
 = x̂
�
E cos φdA =
E⊥ dA
� = E0 (cos φx̂ + sin φŷ)
E
E0 (cos φx̂ + sin φŷ) · dA x̂ = E0 cos φA
More generally we have a curved surface with
� = Ex (x, y, z)x̂ + Ey (x, y, z)ŷ + Ez (x, y, z)ẑ
E
and the integral will be hard to evaluate.
Monday, September 12, 2011
�
Closed Surfaces + Gauss’s Law
• We
are going to be concerned
with the total electric flux
through a closed surface.
•
In advanced integral calculus
E&M, you learn that the
surface integral of the flux can
be related to a volume integral
that depends on the charge in
the volume.
• The
book does not show this;
we will just accept the result:
ΦE =
�
qenclosed
�
�
E · dA =
�0
The result does not depend on the shape of the closed surface!
Monday, September 12, 2011
Closed Surfaces
• There is a charge distribution near the origin.
• The charge furthest from the origin is a distance a away.
• There is a Gaussian sphere of radius b centered at the
origin, with b > a.
•In
which of the following cases does the flux through the
surface definitely not change?
•There
may be more than one right answer.
A. The radius of the sphere decreases to a/2.
B. The magnitude of each charge is doubled.
C. The sphere moves sideways so its surface goes through the origin.
D. The sphere deforms into a cube with each side 2a long centered at
the origin.
E. The charge doubles in magnitude and the sphere doubles in surface
area to compensate.
Monday, September 12, 2011
Closed Surfaces
• There is a charge distribution near the origin.
• The charge furthest from the origin is a distance a away.
• There is a Gaussian sphere of radius b centered at the
origin, with b > a.
•In
which of the following cases does the flux through the
surface definitely not change?
•There
may be more than one right answer.
A. The radius of the sphere decreases to a/2. A.,B.,C.: charge would
usually change, but might
B. The magnitude of each charge is doubled.
not in special cases
C. The sphere moves sideways so its surface goes through the origin.
D. The sphere deforms into a cube with each side 2a long centered at
the origin.
E. The charge doubles in magnitude and the sphere doubles in surface
E. charge definitely changes
area to compensate.
Monday, September 12, 2011
Flux through Closed Surfaces
Monday, September 12, 2011
Closed Surfaces
•
•
Six charges are in a plane.
For simplicity here, treat all
charges as +/-1 μC, ignore the
magnitudes shown.
• The
intersections of 5 closed
surfaces with the plane are shown.
A. S1
B. S2
• Through which surface is the flux greatest?
•There may be more than one right answer
C. S3
D. S4
E. S5
Monday, September 12, 2011
Closed Surfaces
•
•
Six charges are in a plane.
For simplicity here, treat all
charges as +/-1 μC, ignore the
magnitudes shown.
• The
intersections of 5 closed
surfaces with the plane are shown.
A. S1
• Through which surface is the flux greatest?
• There may be more than one right answer.
• q in S1 = 0, q in S2 = +1, q in S3 = +2, q in S4
0, q in S5 = +2 (all in μC).
B. S2
=
C. S3
D. S4
E. S5
Monday, September 12, 2011
See you Thursday
Monday, September 12, 2011