Problem 1:
A sphere of radius R has a uniform volume charge density ρ in addition to a
surface chare density σ = σ0sin2θ. Find the potential
Φ(r) everywhere outside
of the sphere.
Hint:
P0(x) = 1
P1(x) = x
P2(x) = (3x2 - 1)/2
P3(x) = (5x3 - 3x)/2
Solution:
- Concepts:
The principle of superposition, boundary value problems, azimuthal
symmetry
- Reasoning:
The potential outside the sphere is the sum of the potentials of a uniformly
charged sphere and the potential of a sphere with surface charge density σ = σ0sin2θ.
We set the potential at infinity equal to zero.
The potential of the uniformly charged sphere for r > R is
Φ1(r,θ) = (4πε0)-1(4πρR3/3)/r
= ρR3/(3ε0r).
The most general solution inside and outside of a sphere with given surface
charge density and zro volume charge density is
Φ2(r,θ) = ∑n=0∞[Anrn
+ Bn/rn+1]Pn(cosθ).
To find the specific solution we apply boundary conditions.
- Details of the calculation:
σ(R,θ) = σ0 sin2θ =
σ0(1 - cos2θ) = (2σ0/3)(P0
- P2(cosθ)).
Assume
Φ2(r,θ) = B0/r
+ (B2/r3)P2(cosθ)
outside the sphere,
Φ2_in(r,θ) = A0 + (A2r2)P2(cosθ)
inside the sphere.
The symmetry of the situation suggests this form of the solution. If we
can satisfy the boundary conditions, the uniqueness theorem guaranties that
we have found the only solution.
Boundary conditions:
Φ2(R,θ) = Φ2_in(R,θ) --> A0 = B0/R,
A2 = B2/R5.
(E2 - E1)·n2
= σ/ε0.
-∂Φ2/∂r|R + ∂Φ2_in/∂r|R =
σ/ε0.
B0/R2 + (3B2/R4)P2(cosθ)
+ (2A2R)P2(cosθ)
= (2σ0/(3ε0))(1
- P2(cosθ)).
B0/R2 = 2σ0/(3ε0),
(3B2/R4) + (2A2R) = -2σ0/(3ε0).
B0 = 2σ0R2/(3ε0), B2
= -2σ0R4/(15ε0).
Φ2(r,θ) = 2σ0R2/(3ε0r)
- (2σ0R4/(15ε0r3))P2(cosθ).
The potential everywhere outside of the the sphere is
Φ(r,θ) = Φ1(r,θ) + Φ2(r,θ) = (ρR3+ 2σ0R2)/(3ε0r)
- (2σ0R4/(15ε0r3))P2(cosθ)
= (ρR3+ 2σ0R2)/(3ε0r)
+ (σ0R4/(15ε0r3))
- (σ0R4/(5ε0r3))cos2(θ).
This can be rewritten as
Φ(r,θ) =
(ρR3+ 2σ0R2)/(3ε0r) - (2σ0R4/(15ε0r3))
+ (σ0R4/(5ε0r3))sin2(θ).
Problem 2:
A small conducting ring of radius a is located in the xy-plane centered at
the origin. A positive charge Q is place on the ring.
(a) Find the potential Φ on the z-axis at z > a and expand your
expression in powers of a/z.
(b) The potential Φ(r,θ) at a arbitrary points in space with r >
a can be
expanded in terms of Legendre polynomials, Φ(r,θ) = ∑n=0∞[Anrn
+ Bn/rn+1]Pn(cosθ).
Find the expansion coefficients.
Solution:
- Concepts:
Symmetry, Φ(r,θ) = ∑n=0∞[Anrn
+ Bn/rn+1]Pn(cosθ) is the
general solution to Laplace's equation for problems with azimuthal symmetry.
- Reasoning:
Φ(r) = [1/(4πε0)]∫ρ(r')dV'/|r -
r'|.
We can evaluate this integral for r = rk.
For r > a, Φ(r) satisfies Laplace's equation, ∇2Φ(r) =
0. The general solution is
Φ(r,θ) = ∑n=0∞[Anrn + Bn/rn-1]Pn(cosθ).
We can find the coefficients by comparing the general solution to the solution
found by integration on the z-axis.
- Details of the calculation:
(a) Φ(z) = [1/(4πε0)]Q/(a2 + z2)½.
Expanding (a2 + z2)-½ = (1/z)(1 + a2/z2)-½
= (1/z)(1 - ½a2/z2 + ½(3/2)a4/z4
- ½(3/2)(5/2)a6/z6 + ...)
= (1/z)(1 + ∑n=1∞(-1)n(a/z)2n(2n - 1)!!/2n).
Φ(z) = [Q/(4πε0z](1 + ∑n=1∞(-1)n(a/z)2n(2n - 1)!!/2n).
(b) For r > a ∑n=0∞(Bn/rn+1)Pn(cosθ).
Only the even parity terms contribute and all An are zero since Φ = 0
at infinity.
We can therefore write Φ(r,θ) = ∑n=0∞(B2n/r2n+1)P2n(cos(θ)).
On the positive z-axis Φ(z) = ∑n=0∞(B2n/z2n+1),
since Pn(1) = 1 for all n.
Equating the two expressions for Φ(z), we
have
Φ(z) = ∑n=0∞(B2n/z2n+1) =
[Q/(4πε0)](1/z + ∑n=1∞(-1)n(a2n/z2n+1)(2n
-1)!!/2n).
Equating term with equal powers of z we have B0 = Q/(4πε0),
B2n≠0 = [Q/(4πε0)](-1)na2n(2n -1)!!/2n.
Problem 3:
An electron at a distance d = 1 mm is projected parallel to a grounded
perfectly conducting sheet with an energy of 100 electron volts. Let the
grounded conducting sheet lie in the xy-plane and the initial velocity of the
electron point in the x-direction. Let z(0) = d.
(a) Find the z-component vz of the electron's velocity as a
function of its distance z from the sheet.
(b) Find the distance that the electron travels until it hits the plate.
Neglect the force of gravity.
(c) Find the magnitude and direction of a magnetic field parallel to the
surface of the plate and perpendicular to the electron velocity that keeps the
electron from hitting the plate.
Give numerical answers for parts (b) and (c).
Hint: z = dsin2θ may be a useful change of variable. ∫sin2θ dθ
= θ/2 - sin(2θ)/4.
Solution:
- Concepts:
The method of images
- Reasoning:
We calculate the electrostatic force on the electron due to its image charge.
- Details of the calculation:
(a) The force on the electron when it is a distance z from the plane is Fz
= -kqe2/(2z)2.
ax = 0, vx = constant = v0 = (2mE)½
= 5.93*106 m/s.
md2z/dt2 = mdvz/dt = -kqe2/(2z)2. mvzdvz/dt = -¼kqe2(1/z2)(dz/dt).
d/dt(½mvz2) = ¼kqe2d/dt(1/z).
½mvz2 = ¼kqe2(1/z - 1/d).
vz = dz/dt = -[(kqe2/(2m)) (1/z - 1/d)]½.
We can also use the work-kinetic energy theorem to obtain the same result.
If the charge is a distance z from the conducting sheet, it is a distance 2z
from its image. The force on the charge is Fz = -q2/(16πε0z2),
and its potential energy (the negative of the work necessary to remove the
charge from its position to infinity) is U = -q2/(16πε0z).
When a point charge moves from z = d to z = z', its potential energy changes by
ΔU = U(z') - U(d). A point charge released from rest at z = d therefore has
kinetic energy ½mv2 = [q2/(16πε0)][1/z' - 1/d]
at z'.
(b)
tground = ∫0ddz/[(kqe2/(2m))
(1/z - 1/d)]½ = (2md/kqe2)½∫0ddz(z/(d
- z))½.
Let z = dsin2θ, dz = 2dsinθcosθdθ, z½ = d½sinθ,
(d - z)½ = d½cosθ.
∫0ddz(z/(d - z))½ = 2d∫0π/2sin2θdθ
= 2dπ/4.
tground = (π/qe)(md3/(2k))½ =
4.4*10-6 s.
During this time the electron travels 26.2 m in the x-direction.
(c) We need a magnetic field B = -Bj pointing in the negative
y-direction.
Then -qevi×(-B)j = qevBk.
qevB = kqe2/(2d)2, B = kqe/(4vd2)
= 6*10-11 T.
Problem 4:
Assume in a region of space the electric field E lies in the (x,y)
plane and is rotating with frequency ω counterclockwise about the z-axis.
At t = 0, E = E0 i. Assume ω is small and induced
magnetic fields can be neglected.
(a) How could you produce such an electric field?
(b) Find general solution for x(t), y(t), vx(t) and vy(t)
for a particle with mass m and charge q located in that region of space?
(c) Is it possible for the particle to stay confined in that region of
space? What initial conditions are required?
Solution:
- Concepts:
The electric force F = qE
- Reasoning:
The rotating field can be viewed as a superposition of two oscillating electric fields.
- Details of the calculation:
(a)
E = E0cos(ωt) i + E0sin(ωt)
j. Large (compare to the dimensions of the region) parallel plate
capacitors with sinusoidally varying voltages across the plates could produce
such a field.
(b) d2x/dt2 = (qE0/m)cos(ωt).
d2y/dt2
= (qE0/m)sin(ωt).
Let ζ = x + iy.
The d2ζ /dt2 = (qE0/m)exp(iωt).
Inhomogeneous solution:
Try ζ = ζ0exp(iωt), -ω2ζ0 = (qE0/m) ζ0 =
-qE0/(mω2).
Homogeneous solution:
ζ = A +
Bt, A, B = arbitrary complex constants.
General solution:
ζ = A
+ Bt - [qE0/(mω2)]exp(iωt).
x(t) = |A|cos(φA) + |B|t cos(φB) - [qE0/(mω2)]cos(ωt).
y(t) = |A|sin(φA) + |B|t sin(φB) - [qE0/(mω2)]sin(ωt).
vx(t) = |B| cos(φB) + [qE0/(mω)]sin(ωt).
vy(t) = |B| sin(φB) - [qE0/(mω)]cos(ωt).
(c)
For the particle to stay confined to the region, the constant B has to be zero.
This implies vx(0) = 0, vy(0) = -[qE0/(mω)].
Problem 5:
The displacement vector from electric dipole p1 to dipole p2 is
r.
(a) Calculate the electric potential energy W;.
(b) Calculate the force F21 that p1 exerts on
p2.
(c) Calculate the torque τ12 that p1 exerts on
p2.
(d) Let p1 = (10-9 Cm)k be located at
the origin and p2 = (10-9 Cm)i at r =
(3 m)i + (4 m)k.
Provide numeric answers for F21 and τ12.
Solution:
- Concepts:
The dipole field, a dipole in an external field
- Reasoning:
We have one dipole in the external field of another dipole.
The energy of a dipole in an external field is U = -p∙E.
- Details of the calculation:
(a)
Consider dipole p2 in the external field of dipole
p1. Choose your coordinate system such that
p1 is at the origin
and p2 is at r.
Field of p1 at r:
E1(r) = (1/(4πε0))[3(p1∙r)r/r5
- p1/r3].
Energy of p2 in this field: U(r) = -p2∙E1(r)
=
(1/(4πε0))[p1∙p2/r3
- 3(p1∙r)(p2∙r)/r5].
(b) F21(r) = -∇U(r) = -(1/(4πε0))[∇(p1∙p2/r3)
- 3∇((p1∙r)(p2∙r)/r5)].
p1 and p2 are constant, ∇(p1∙p2)
= 0, ∇(pi∙r)= (pi∙∇)r
= pi.
F21(r) = -(1/(4πε0))[(p1∙p2)∇(1/r3)
- 3[(p1∙r)(p2∙r)∇(1/r5)
+ p2(p1∙r)/r5 + p1(p2∙r)/r5]].
∇(1/r3) = -3r/r5. ∇(1/r5)
= -5r/r7.
F21(r) = (3/(4πε0r5))[(p1∙p2)r
+ p2(p1∙r) + p1(p2∙r)] - 5(p1∙r)(p2∙r)(r/r2)].
(c) τ12(r) = p2×E1(r)
= (1/(4πε0))[3(p1∙r)(p2×r)/r5
- (p2×p1)/r3].
(d) In SI units: p1∙p2 =
0. p1∙r = 4*10-9. p2∙r
= 3*10-9. p2×r = -4*10-9j.
p2×p1 = -10-18j.
F21 = (3/(4πε055))[(4*10-18 C2m3)i +
(3*10-18 C2m3)k] - 12*10-18C2m3(3i/5
+ 4k/5))]
= -(3/(4πε055))[(3.2*10-18 C2m3)i +
(6.6*10-18 C2m3)k].
= -[(2.8*10-11 N)i + (5.7*10-11 N)k].
τ12 = (1/(4πε055))[-36*10-18j
+ 25*10-18j]Nm = (1/(4πε055))[-11*10-18j]
Nm = -3.17*10-11 Nm j.