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Unformatted text preview: • Physics 227: Lecture 3
Electric Field Calculations, Symmetry,
Charged Particle Motion, Flux, Gauss’s Law
Lecture 2 review: If you know anyone who has not managed to register,
some more spaces have opened up. Please have them
r2 contact Stacey Jacbs ASAP. •
• 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
ˆ
2 ri →
4π0 ri
i qi
ˆ
2 ri →
4π0 ri ρ( )
x
ˆ
∆
dx
4π0  − 2
xx 1ˆ
dq
∆
4π0 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
ˆ
2 ri →
4π0 ri x d ρ( )
x
ˆ
∆
dx
4π0  − 2
xx
3 What is the electric ﬁeld along the x axis?
For x > d: For 0 < x < d: Monday, September 12, 2011 q2
q1
x+
ˆ
x
ˆ
E=
4π0 x − d2
4π0 x + d2
−q1
q2
E=
x+
ˆ
x
ˆ
2
2
4π0 x − d
4π0 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 qi
ˆ
2 ri →
4π0 ri ρ( )
x
ˆ
∆
dx
4π0  − 2
xx
3 •
x What is the electric ﬁeld along
the y axis? • You can see the vertical
components will cancel. −2q1
d
E=
x×
ˆ
4π0 (y 2 + d2 )
y 2 + d2
−2q1 d
E=
x
ˆ
4π0 (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
x faster than the ﬁeld
ˆ
4π0 (y 2 + d2 )3/2
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=
x faster than the ﬁeld
ˆ
told enough so far to know
4π0 (y 2 + d2 )3/2
of a point charge
this answer is wrong
Monday, September 12, 2011 Electric Field of a Ring of Charge Example 21.10 •
• x
E ( ) =
i qi
ˆ
2 ri →
4π0 ri 1ˆ
dq
∆
4π0 r2 You can see the vertical
components will cancel. •
What is the electric ﬁeld
along the x axis? 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 Eﬁeld 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 Eﬁeld, 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. 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 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 Ushaped
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
Eﬁeld lines ΦE = ˆ
Surface: A
unit vector
⊥ to surface try ΦE =
E · dA = ˆˆ
A=x E cos φdA = E⊥ dA
E = E0 (cos φx + sin φy )
ˆ
ˆ E0 (cos φx + sin φy ) · dA x = E0 cos φA
ˆ
ˆ
ˆ More generally we have a curved surface with
E = Ex (x, y, z )ˆ + Ey (x, y, z )ˆ + Ez (x, y, z )ˆ
x
y
z
and the integral will be hard to evaluate.
Monday, September 12, 2011 Closed Surfaces + Gauss’s Law
• We are going to be concerned
w ith 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 = · dA = qenclosed
E
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 ...
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This note was uploaded on 09/27/2011 for the course PHYSICS 750:227 taught by Professor Ronaldgilman during the Fall '11 term at Rutgers.
 Fall '11
 RonaldGilman
 Physics, Charge

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