Assignment 9, solutions

Problem 1:

(a)  A positive charge Q is spread over an semicircular arc with radius R as shown.
(1)  What is the work required to bring in a charge -q from infinity to the center of the arc?
(2)  Calculate the magnitude and direction of the force on a charge –q at the center of the arc.

(b)  The potential of a uniformly charged spherical shell of radius R centered at the origin is
V(r) = q/(4πε0r) r ≥ R,   V(r) = q/(4πε0R) r < R, 
where q denotes the total charge of the sphere.
Calculate the energy that it requires to deposit a charge Q on an initially neutral conducting spherical shell with radius R.  Use two different approaches to come to the result.
(3)  Calculate the energy by incrementally adding a charge dq to the sphere.
(4)  Obtain the energy by considering the resulting electric field of the spherical shell.

Solution:

bulletConcepts:
Electrostatic potential energy and work
bulletReasoning:
The electric field is a conservative field, the work does not depend on the path.
bulletDetails of the calculation:
(1)  The potential (work per unit charge) due to the charge Q at the origin is 
V = kQ/R,  k = 1/(4πε0), SI units.
W = -kqQ/R is the work required to bring in a charge -q from infinity to the center of the arc.
(2)  The force will be F = Fj, pointing in the y-direction from symmetry.
F = 2kq∫0π/2λRdθ sinθ/R2 = (2kqλ/R)∫0π/2sinθ dθ = (2kqλ/R), with λ = Q/(πR).
F = 2kqQ/(πR2).
(3)  When a charge q has already been added to the sphere, the additional work needed to add another charge dq is kqdq/R.  The total work necessary to deposit a charge Q on the shell is
W = (k/R)∫0Qqdq = ˝kQ2/R.  This is the energy stored in the charge distribution.
(4)  E = kQ/r2 outside the shell, E = 0 inside the shell.
The energy density in the electric field is u(r) = (ε0/2)E(r)2.  
The total energy stored in the field is
4π∫Rr2dr u(r) = (4πε0/2) ∫Rr2dr k2Q2/r4 = [Q2/(8πε0)] ∫Rdr/r2 = Q2/(8πε0R) = ˝kQ2/R.
In electrostatics, viewing the energy as being stored in the charge distribution or viewing it as being stored in the electric field leads to the same results.

Problem 2:

A model of the hydrogen atom was proposed before the advent of quantum mechanics, which consists of a single electron of mass m and an immobile uniform spherical distribution of positive charge with radius R.  Assume that the positive charge interacts with the electron via the usual Coulomb interaction but otherwise does not offer any resistance to the motion of the electron.
(a)  Explain why the electron’s equilibrium position is at the center of the positive charge.
(b)  Show that the electron will undergo simple harmonic motion if it is displaced a distance d < R away from the center of the positive charge.  Calculate its frequency of oscillation.
(c)  How big would the atom need to be in order to emit red light with a frequency of 4.57*1014 Hz?  Compare your answer with the radius of the hydrogen atom.
(d)  If the electron is displaced a distance d > R from the center, will it oscillate in position?  Will it undergo simple harmonic motion?  Explain!

Solution:

bulletConcepts:
Gauss' law
bulletReasoning:
The field due to a spherically symmetric charge distribution can be found from Gauss’ law.
bulletDetails of the calculation:
The charge distribution is spherically symmetric and so the electric field is  E(r) = E(r) r/r.
Let the center of the positive charge be the origin and consider a Gaussian surface defined to be a sphere or radius r at the same origin.
The total flux is Φ = ∫E∙dA = 4πr2E(r) = Qinside0.
If 0 < r < R, then Qinside = 4πρr3/3, with ρ = 3qe/4πr3.
In this region E(r) = ρr/(3ε0).
If r > R, the electric field is the same as expected for a point charge, E(r) = qe/4π ε0r2.
We can now easily answer the questions.
(a)  The force acting on the electron is F(r) = -qeρr/(3ε0), pointing towards the origin.  Therefore the only point where the sum of all forces is zero is the origin.
(b)  The force is a restoring force and is proportional to the displacement of the electron from r = 0.  The resulting motion will be simple harmonic motion.  
The “spring constant” associated with this motion is k = qeρ/(3ε0) = qe2/(4πε0R3).
The frequency of the oscillations is therefore
f = (2π)-1(k/m)1/2 = (2π)-1(qe2/(4πε0mR3))1/2.
(c)  R = [qe2/(16π3ε0mf2)]1/3.  For f = 4.57*1014 Hz we need R = 3.13*10-10 m. 
(d)  For d > R the force is still restoring.  Therefore the electron will undergo oscillatory motion.  The magnitude of the force is proportional to 1/r2, however, and therefore the motion will not be simple harmonic motion.

Problem 3:

A charge array consists of two charges, each of magnitude +q, located on the z-axis at (0, 0, ±a).
(a)  Find the potential V(0,0,z) at an arbitrary point z > a; then expand V(0, 0,z) in a power series in z.
(b)  Using this series as a “boundary condition”, find the potential V(r,θ,φ) at an arbitrary location (r,θ,φ) with r > a.  An infinite series is acceptable.
(c)  Characterize the first three terms in 1/rn.

Solution:

bulletConcepts:
The electrostatic potential
bulletReasoning:
In regions with charge density ρ = 0 we have ∇2V = 0.  So in the region r > a we have
2V = 0.  For a problem with azimuthal  symmetry, the most general solution for ∇2V = 0 is
V(r,θ) = Σn=0(Anrn + Bnr-(n+1))Pn(cosθ).
On the positive z-axis Pn(cosθ) = 1 and V(z) = Σn=0(Anzn + Bnz-(n+1)).
If we know the potential on the z-axis, we can solve for the expansion coefficients by equating term of equal power in z.
bulletDetails of the calculation:
(a)  V(z) = kq/(z - a) + kq/(z + a) = (kq/z)[(1 - a/z)-1 + (1 + a/z)-1].
Assume a/z < 1.
For x < 1 we have (1 ± x)-1 = 1 ∓ x + x2 ∓ x3 + … .
V(z) = (2kq/z)(1 + (a/z)2 + (a/z) 4 + …) = 2kq(1/z + a2/z3 + a4/z 5 + …).
(b)  V(z) = 2kq(1/z + a2/z3 + a4/z 5 + …) = Σn=0(Anzn + Bnz-(n+1)).
An = 0 for all n.  Bn = 0 for n = odd,  Bn = 2kqan for n = even.
V(r,θ) = 2kqΣn=even anr-(n+1)Pn(cosθ).
(c)  First term: Monopole term  ∝  1/r = 2kq/r. 
Second term:  Dipole term  ∝  1/r2 = 0.
Third term: Quadrupole term  ∝  1/r3 = 2kqa2P2(cosθ)/r3.

Problem 4:

Compute the force of attraction between a neutral metallic sphere of radius a and a point charge q positioned a distance r from the center of the sphere, where r > a.
How does the force of attraction behave as a function of r if r >> a?

Solution:

bulletConcepts:
The method of images
bulletReasoning:
Orient the coordinate system so that the sphere is centered at the origin and the point charge q is located at  z = r on the z-axis.
Placing an image charge q' = -qa/r at z' = a2/r  on the z-axis makes the surface of the sphere an equipotential with Φ = 0.  Adding q'' = qa/r at the center of the sphere changes the total charge on the sphere and makes it neutral while keeping its surface an equipotential.
bulletDetails of the calculation:
The electrostatic force between the charge q and the grounded sphere has the same magnitude and direction as the electrostatic force between the charge q and the image charges q' and q''.  It is attractive since q' is closer to q than q''.  The magnitude of the force is less than the force between the charge and the grounded sphere.
|F| = (kq2a/r)[1/(r – a2/r)2 – 1/r2] =  (kq2a/r3)(2a2/r2 – a4/r4)/(1 – a2/r2)2
=  kq2a3(2r2 – a2)/(r2 – a2)2r3.
Here k = (4πε0).
If r >> a we have |F| ≈ (kq2a/r3)(2a2/r2) = (2kq2a3/r5).

Problem 5:

A spherical charge distribution is given by
ρ = ρ0(1 - r/a),    r < a,
ρ = 0,    r > a.
(a)  Calculate the total charge Q.
(b)  Find the electric field and potential for r > a.
(c)  Find the electric field and potential for r < a.
(d)  Find the electrostatic energy of this charge distribution.

Solution:

bulletConcepts:
Gauss' law
bulletReasoning:
The field due to a spherically symmetric charge distribution can be found from Gauss’ law.
bulletHow do they apply?
(a)  Q = 4πρ00a r2dr(1 - r/a) = 4πρ0[(a3/3) - (a3/4)] = 4πρ0a3/12.
(b)  E = (1/(4πε0))Q/r2, radially outward for positive ρ0.
E = (1/ε00a3/(12r2).
Φ = (1/(4πε0))Q/r = (1/ε00a3/(12r).
(c)  E = (1/(4πε0)) Qinside/r2, radially outward for positive ρ0.
Qinside = Q = 4πρ00r r’2dr’(1  -r’/a)  = 4πρ0[(r3/3) - (r4/(4a))].
E = (1/ε00[(r/3)  -(r2/(4a))].
Φ = Φ(a) + ∫raE(r)dr = Φ(a) + (ρ00)∫ra[(r/3) - (r2/(4a))]dr
= Φ(a) + (ρ00)[(r2/6) - (r3/(12a))]ra
= (ρ00)[a2/12 + a2/6 - a2/12  - r2/6 + r3/(12a)]
= (ρ00)[a2/6  - r2/6 + r3/(12a)].
(d) U = (ε0/2)∫all_space E∙E dV
U = (ρ02/2ε0)∫0a[(r2/9) - (r3/(6a)) + (r4/(16a2))]4πr2dr
+ (ρ02a6/288ε0)∫a(1/r4)4πr2dr
= (ρ02a50)*0.0648.
or
U = (1/2)∫0aρ(r)Φ(r)dV = (4π/2)∫raρ(r)Φ(r)r2dr
= (2πρ020)∫0a(1 - r/a)[a2/6  - r2/6 + r3/(12a)]r2dr
= (ρ02a50)*0.0648.