Download Physics 227: Lecture 6 Dipoles, Calculating Potential Energy or

Physics 227: Lecture 6
Dipoles, Calculating Potential Energy
or Potential, Equipotential Lines
•
Lecture 5 review:
•
•
•
The electric field vanishes inside a spherical shell of charge.
Conductors and surface charge.
Conservative forces, fields, potential energy, work, potentials:
F = -∇U, U = -∫F.dx ➭ UC = kQq/r.
Thursday, September 22, 2011
The Dipole
•
Put a pair of charges, ±q, a
fixed distance d apart in a
constant E field.
• Why
are dipoles important?
Although molecules are typically
neutral, there are some, like H2O,
that have a static dipole moment.
• The
Picture copied out of Wikipedia
Thursday, September 22, 2011
dipole moment of water is
≈6x10-30 C.m. If we take one
electron for q, then d = p/q ≈ 40
pm, nearly the size of a hydrogen
atom (r ≈ 40 pm). The electrons
move away from the H towards
the O.
The Dipole
•
Put a pair of charges, ±q, a
fixed distance d apart in a
constant E field.
• The
total force from the
external field is 0... but it
generates a torque that causes
the dipole to rotate:
τ = 2qE(d/2) sinϕ = qEd sinϕ.
• The
structure of the dipole comes in as a product qd, which
we define as the dipole moment: p = qd.
• Thus: τ = pE sinΦ, or in vector form: τ = p x E
• You can see that the work done by the electric
field is
dW = 2 q E (d/2) sinΦ (-dΦ) = -pE sinΦ dΦ = pE d(cosΦ).
• Thus:
Thursday, September 22, 2011
dU = -dW = -pE d(cosΦ), or U=-pE cosΦ = -p.E
The Dipole
•
Put a pair of charges, ±q, a
fixed distance d apart in a
constant E field.
• The
total force from the
external field is 0... because the
field is constant. If the field is not
uniform, then we would expect
that usually the force is non-zero.
Thursday, September 22, 2011
Calculating Potential / Potential Energy
•
Generally there are two ways to do this.
•
The electric field E is specified, and you do a line
integral of E.dl.
•
The charge distribution is specified and you do a
volume integral of kρ/r or a sum over charges of
kqi/r.
•
Thursday, September 22, 2011
Potential is a scalar, not a vector quantity, and
the sums and integrals are similar to those for
ˆ 2→r
determining the electric field: r/r
Single Point Charge Potential,
Potential Energy of Two charges
•
•
V = kQ/r
•
The standard
convenient choice is V
= 0 at r = ∞. There is
no point in adding a
constant to this.
U = kQq/r
•
Again we set the
potential energy to 0
at r = ∞.
Thursday, September 22, 2011
Potential Energy for Two Charges
• The
potential energy has the same magnitude, but
opposite sign, depending on whether the charges have
the same or opposite signs.
Thursday, September 22, 2011
Potential from a ring of charge
•
On the axis, we can get the
potential from a ring of charge by
integrating:
V(r) = ∫k dq / r = kq/r.
• The
formula is the same result as
for a point charge, because on the
axis all points on the loop are equally
far away and we use the variable r.
•
Note rminimum = a; r does not go to 0.
• With
a point charge at O, we have on the x-axis V(x) = kq/x, the
same formula as V(r) except for a change of variables.
• Whereas
for the ring of charge we have V(x) = kq/(x2+a2)1/2.
We have the same formula and different limits on r,
or the same variable x and different formulas.
Thursday, September 22, 2011
Potential iClicker
For the three charge distributions
shown, which of the choices shown
for the orderings of the potential
at the point X is correct?
You may assume that the charge
density per unit length along the
circular arcs is the same in all
cases. You may assume the charge
on the ring is positive.
Thursday, September 22, 2011
A. VA = VB = VC.
B. VA > VB > VC.
C. VA > VB = VC.
D. VA = VB > VC.
E. none of the others.
Potential iClicker
As per the ring of charge
derivation, the field for each is
kqi/r, since all the charge is r
away from the central x, but qB =
qC while qA is twice as large. So
answer C.
Thursday, September 22, 2011
A. VA = VB = VC.
B. VA > VB > VC.
C. VA > VB = VC.
D. VA = VB > VC.
E. none of the others.
Another Potential iClicker
For the three charge distributions
shown, which of the choices shown
for the orderings of the potential
at the point X is correct?
But now assume that the total
charge in each of the 3 loops is
the same - B and C have the same
charge density, and A has half the
charge density of B and C. You
may assume the charge on the
ring is positive.
Thursday, September 22, 2011
A. VA = VB = VC.
B. VA > VB > VC.
C. VA > VB = VC.
D. VA = VB > VC.
E. none of the others.
Another Potential iClicker
A. VA = VB = VC.
Since the total charge is not the
same for each ring, the potential
is the same in each case as well.
B. VA > VB > VC.
C. VA > VB = VC.
D. VA = VB > VC.
E. none of the others.
Thursday, September 22, 2011
Potential vs. Field for a
Charged Spherical Shell
•
Outside the spherical shell, we
know E(r) = kq/r2, and the
potential is V(r) = kq/r.
• What
happens inside the
spherical shell?
• The
electric field is 0 inside
the spherical shell.
•If
E=0, then ΔV = -∫E.dl = 0, so
the potential is constant.
Thursday, September 22, 2011
Potential Generated by Two
Fixed Charges
y
• When
r1
we have a charge > 0
and one < 0, the potential is V(r)
= kq1/r1 - kq2/r2.
r2
• The
+q1
Thursday, September 22, 2011
-q2
x
potential is 0 wherever
q1/r1 = q2/r2, or r1 = (q1/q2)r2.
•
If q1 = q2, this defines the y
axis. Otherwise we have a
closed curve around the smaller
magnitude charge.
Potential from an infinite
Line of charge
•
Integrating kdq/r looks harder,
vs. integrating (λ/2πε0r)dr which
looks easier:
∆V = −
• As
• The
�
a
b
λ
r̂ · d�r
2π�0 r
∫dr/r → ln(r), we obtain:
� r0 �
λ
ln
∆V =
2π�0
r
potential decreases as r
increases. It goes to 0 at some
reference radius r0, and then
becomes negative. We cannot
choose r0 = ∞, as then V = ln(∞)
= ∞ everywhere.
Thursday, September 22, 2011
Potential from an infinite
cylinder of charge
•
For an infinite cylinder of
charge, the potential is the
same outside the cylinder. Using
λ = πr2ρ, we have again:
•
� r0 �
λ
ln
∆V =
2π�0
r
It is convenient to choose r0 =
R, the radius of the cylinder.
• What
happens inside the
cylinder?
• E = (ρ/2ε0)r, so V= -(ρ/4ε0)r2 + arbitrary constant.
2.
To
match
V=0
at
r=R,
the
constant
is
(ρ/4ε
)R
0
•
Thursday, September 22, 2011
Potential for a Parallel PLate Capacitor /
infinite Planes of charge
+
+
+
+
+
+
+
+
+
+
-
-
-
Charge density: ±σ C/m2 on each conducting plane
E
+
+
• We
assume the plate area A is square and large compared to the
plate separation d (√A = l >> d). That is, we ignore fringe fields and
treat the plates as infinite in extent, and the E field as constant.
• Drawing a Gaussian surface, we see EAG = σAG/ε0 → E = σ/ε0.
• The voltage between the plates is V = ∫E.dl = Ed = σd/ε0 = Qd/Aε0.
• We will come back to the energy stored in a capacitor next week.
Thursday, September 22, 2011
Algebraic Problem Example
V = 4x + 3y − 2z
2
3
This potential pulled out of “thin air”.
� d
� 2
d
d
3
�
E = −∇V = −
x̂ +
ŷ + ẑ (4x + 3y − 2z )
dx
dy
dz
2
�
E = −8xx̂ − 3ŷ + 6z ẑ
V = 5xy
�
E = −5yx̂ − 5xŷ
This potential pulled out of “thin air”.
Thursday, September 22, 2011
Graphical Problem Example
V(x) is shown. Which plot shows the correct E(x)?
A.
B.
C.
D. None of them.
Thursday, September 22, 2011
Spherical Conductor with Cavity
There is a charge +q at the center of the cavity.
+q
What does the electrical potential look like?
V
V
V
C.
r
Thursday, September 22, 2011
V
A.
r
D.
B.
r
V
E.
r
r
F. A-E are all nonsense.
E ≈ 1/r2
Spherical Conductor with Cavity
0
There is a charge +q at the center of the cavity.
1/r2
+q
What does the electrical potential look like?
V
V
A.
B.
V ≈ 1/r constant 1/r
V
V
C.
r
Thursday, September 22, 2011
r
D.
r
V
E.
r
r
F. A-E are all nonsense.
Equipotential Lines
•
•
Topographic maps shows
lines of constant
elevation = lines of
constant gravitational
potential - let’s not even
discuss any potential
issues due to the
earth’s non-sphericity
and rotation
When the lines are
closer together, the
slope is greater.
Equipotential plots in electrostatics play the same role as the
lines of constant elevation in topographic maps.
Thursday, September 22, 2011
Equipotential Lines
•
•
•
•
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The familiar field lines are shown in red.
The equipotential lines are shown in blue.
Field lines have a direction, whereas equipotential lines do not.
Field lines do not touch or cross, but equipotential lines can, where E=0.
Each is perpendicular to the other. Why? E = -∇V and the direction of
the greatest change in V is the perpendicular to the V=constant line.
Thursday, September 22, 2011
Equipotential Lines
•
A puzzle: How can equipotential lines cross if...
•
•
The electric field is a unique vector at each point in space, and
E = -∇V so the field line is the perpendicular to the equipotential line?
Thursday, September 22, 2011
Thank you.
On Monday, Sep 26 I will be traveling.
Prof. Cizewski will be giving the lecture.
On Thursday Sep 29 I will still be away.
Prof. Chandra will give the lecture.
See you Monday Oct 3.
Thursday, September 22, 2011