physics 227: lecture 6 dipoles, calculating potential ... · pdf filephysics 227: lecture 6...

Physics 227: Lecture 6Dipoles, 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 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.

Picture copied out of Wikipedia


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: 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.


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.

• Potential is a scalar, not a vector quantity, and the sums and integrals are similar to those for determining the electric field: r/r2 → rˆ


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 = ∞.


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.


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.


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.

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.

A. VA = VB = VC.

B. VA > VB > VC.

C. VA > VB = VC.

D. VA = VB > VC.



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.

A. VA = VB = VC.

B. VA > VB > VC.

C. VA > VB = VC.

D. VA = VB > VC.



Another Potential iClicker

Since the total charge is not the same for each ring, the potential is the same in each case as well.

A. VA = VB = VC.

B. VA > VB > VC.

C. VA > VB = VC.

D. VA = VB > VC.



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.


Potential Generated by Two Fixed Charges

• When we have a charge > 0 and one < 0, the potential is V(r) = kq1/r1 - kq2/r2.• The 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.

+q1 -q2

r2r1

y

x


Potential from an infinite Line of charge

• Integrating kdq/r looks harder, vs. integrating (λ/2πε0r)dr which looks easier:

∆V = −� b

a

λ

2π�0rr · d�r

• As ∫dr/r → ln(r), we obtain:

∆V =λ

2π�0ln

�r0

r

�

• The 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.


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:

∆V =λ

2π�0ln

�r0

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.


Potential for a Parallel PLate Capacitor / infinite Planes of charge

• 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.

Charge density: ±σ C/m2 on each conducting plane

+ ++ ++ ++ ++ ++ +

- -- -- -- -- -- -E


Algebraic Problem Example

V = 4x2 + 3y − 2z3

V = 5xy

This potential pulled out of “thin air”.

This potential pulled out of “thin air”.

�E = −∇V = −� d

dxx +

d

dyy +

d

dzz�(4x2 + 3y − 2z3)

�E = −8xx− 3y + 6z2z

�E = −5yx− 5xyThursday, September 22, 2011

Graphical Problem Example

V(x) is shown. Which plot shows the correct E(x)?

A.

B.

C.

D. None of them.


Spherical Conductor with Cavity

What does the electrical potential look like?

There is a charge +q at the center of the cavity.+q

r

V

r

V

r

V

r

V

A. B.

C. D.

r

V E.

F. A-E are all nonsense.


Spherical Conductor with Cavity

What does the electrical potential look like?

There is a charge +q at the center of the cavity.+q

r

V

r

V

r

V

r

V

A. B.

C. D.

r

V E.

F. A-E are all nonsense.

1/r1/r constantV ≈

E ≈ 1/r2

1/r2

0


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.


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.


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?


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.


physics 227: lecture 6 dipoles, calculating potential ... · pdf filephysics 227: lecture 6...

Documents