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Unformatted text preview: 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
≈6x1030 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 nonzero. 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
determining the electric ﬁeld: r/r2 → 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
o pposite 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 xaxis 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
r · d
ˆr
2π0 r ∫dr/r → ln(r), we obtain: r0
λ
ln
∆V =
2π0
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 =
2π0
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.
To match V=0 at r=R, the constant is (ρ/4ε0)R2.
•
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
d
d 2
3
= −∇V = −
E
x+
ˆ
y + z (4x + 3y − 2z )
ˆ
ˆ
dx
dy
dz = −8xx − 3ˆ + 6z 2 z
E
ˆ
y
ˆ V = 5xy
= −5y x − 5xy
E
ˆ
ˆ 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. AE 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. AE 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 nonsphericity
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 •
•
•
•
• 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 ...
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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, Energy, Force, Potential Energy, Work

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