Problem 1:
Consider a conducting (i.e.
metallic) region defined by two half-planes connected to the ground (i.e. at
zero potential Φ), as shown in the figure. These planes are perpendicular to
one another i.e. one is defined by x = 0 and y > 0, and the other by y = 0 and x
> 0. A charge q is located are position P = (a, a, 0) as shown. The region
where the charge resides is in the vacuum, and it is called here the first
quadrant.
Note that this is a 3D problem, not just 2D.
(a) Using the method of images find the scalar potential
Φ(r) in the first quadrant
at an arbitrary point r = (x, y, z).
(b) Verify that the potential you found in (a) cancels in the two half-planes.
(c) Find the force on the charge q induced by the planes.
Solution:
- Concepts:
Method of images
- Reasoning:
We need an arrangement of charges
such that the potentials add
to zero at the surfaces of the planes. Add three image charges as shown
in the figure. The electric potential and the electric field in the region
between the planes is the same as that produced by the four charges.
- Details of the calculation:
(a) Φ(r) = keq[1/((x
- a)2 + (y - a)2 + z2) + 1/((x + a)2
+ (y + a)2 + z2)
- 1/((x + a)2 + (y - a)2 + z2) - 1/((x - a)2
+ (y + a)2 + z2)].
(b) If x = 0 and y > 0, we have
Φ(r) = keq[1/(a2 + (y - a)2 + z2) +
1/(a2 + (y + a)2 + z2)
- 1/(a2 + (y - a)2 + z2) - 1/(a2 +
(y + a)2 + z2)] = 0.
If y = 0 and x > 0, we have
Φ(r) = keq[1/((x - a)2 + a2 + z2) +
1/((x + a)2 + a2 + z2)
- 1/((x + a)2 + a2 + z2) - 1/((x - a)2
+ a2 + z2)] = 0.
(c) The force on q is directed along the diagonal from the upper right to the
lower left corner and has magnitude
F = 2keq2cos45o/(4a2) - keq2/8a2
= [keq2/(4a2)][√2 - sa½].
Or F = -ex keq2/(4a2)
- ey keq2/(4a2) + ((ex
+ ey)/√2) keq2/(8a2).
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:
For this problem, consider only the Coulomb interaction,
i.e. neglect gravity, magnetism, and radiation.
A point charge of mass m = 10-3 kg and charge q
= 10-6 C and an insulated, conducting spherical shell of mass M = 10-1
kg and radius R = 0.01 m orbit their common center of mass in circular orbits.
The distance between the center of the sphere and the point charge is d = 0.1
m. The spherical shell spins, so that the line from the center of the sphere to the
point charge always passes through the same point on the shell.
Find the total angular momentum of the system about the
center of mass.
(Give a numerical answer!)
Solution:
- Concepts:
Method of images
- Reasoning:
Assume that at some time t the point charge q is located on
the z axis at z = d. Place an image charge q' = -Rq/d on the z-axis at z' = R2/d.
This will keep the sphere at zero potential. The total charge Q = on the sphere
is zero. Place q'' = Rq/d at the center of the spherical shell to
keep it neutral.
- Details of the calculation:
The force on q is Fz = kq2(R/d)[1/d2
- 1/(d - R2/d)2] .
Fz = (kq2/R2)[(1/x3
- x/(x2
- 1)2] = (kq2/R2)(1 - 2x2)/(x3/(x2
- 1)2).
with x = d/R = 10.
Fz is negative, it is an attractive force.
Let μ = mM/(m + M) = 9.9*10-4 kg
be the reduced mass of the system.
The problem of the relative motion of two interacting masses m
and M can be solved by solving for the motion
of one fictitious particle of reduced mass μ
in a central field.
μv2/d = μdω2
= μRxω2
= (kq2/R2)(2x2 - 1)/(x3(x2
- 1)2).
ω2 =
(kq2/(R3μ))(2x2
- 1)/(x4(x2 - 1)2).
ω2 =
[9*109*10-12/(10-6*9.9*10-4)](2*102
- 1)/(104(102 - 1)2) s-2.
ω = 4.3/s.
L = Ishellω + md2ω, where Ishell
= 2MR2/3.
2MR2/3 + md2 = 2*10-1*10-4/3 +
10-3*10-2 = 5*10-5/3.
L = 7.2*10-5 kgm2/s.
Problem 4:
Two conducting cones (0 < θ1
< θ2 < π/2) of infinite extend are separated by an
infinitesimal gap at r = 0.
Let V(θ1) = 0 and V(θ2) = V0. Find V
between the cones.
Laplacian in spherical coordinates:
∫dx/sinx = ln(tan(x/2))
Solution:
- Concepts:
Laplace's equation in spherical coordinates
- Reasoning:
Since the cones are infinite, V can only depend on θ.
(Rotational symmetry --> no φ dependence
System is unchanged if we change the scale --> no r dependence)
∇2V = 0 becomes
(1/(r2sinθ))∂/∂θ(sinθ ∂V/∂θ) = 0.
Since r = 0 is excluded due to the insulating gap, we can write
∂/∂θ(sinθ ∂V/∂θ)
= 0.
- Details of the calculation:
Integrating once gives sinθ ∂V/∂θ = A, or ∂V/∂θ = A/sinθ.
Integrating again gives V = A ln(tan(θ/2)) + B.
Boundary conditions:
V(θ1) = 0 --> B = -A ln(tan(θ1/2)).
V(θ2) = V0 --> A ln(tan(θ2/2)) - A
ln(tan(θ1/2)) = A ln[tan(θ2/2)/tan(θ1/2)] = V0.
A = V0/ln[tan(θ2/2)/tan(θ1/2)].
V(θ) = A ln[tan(θ/2)/tan(θ1/2)] = V0 ln[tan(θ/2)/tan(θ1/2)]/ln[tan(θ2/2)/tan(θ1/2)].