Review problems:

Problem 1:

A satellite has a polar orbit about the earth with a radius of 5.49*106 m above the earth's surface.  On one orbit it passes over Greenwich at 0o longitude and 51.48o latitude. Approximately where will it pass over exactly one orbit later?  The radius of the earth is RE = 6.38*106 m and its mass is 5.98*2024 kg.

Solution:

bulletConcepts:
Kepler's third law
bulletReasoning:
mv2/r  = GMm/r2.  v2  = GM/r.  (2πr/T)2 = GM/r. 
T2 = (4π2/(GM))r3,  Kepler's third law.
bulletDetails of the calculation:
T2 = (4π2/(6.67*10-11 * 5.98*1024))(6.38*106 + 5.49*106)3 = 1.66*108 s2.
T = 1.29*104 s.
The earth rotates through 360o in 24 hours and therefore through 53.6o in one period T.
One orbit later the satellite will pass over a place with the same latitude and 53.6o longitude, west of Greenwich.

Problem 2:

A static charge distribution produces a spherically symmetric, radial electric field E  = A exp(-br)/r2(r/r), where A and b > 0 are constants.
(a)  What is the volume charge density ρ(r)?
(b)  What is the total charge?

Solution:

bulletConcepts:
Gauss' law
bulletReasoning:
E = ρ/ε0.  For the given field, the expression is not well defined at r = 0.
We can handle the divergence at r = 0 by using∙(r/r3) = 4πδ(r).
We can also use the integral form of Gauss' law.
bulletDetails of the calculation:
E = A exp(-br)∙(r/r3) + (r/r3)∙(A exp(-br)) = ρ(r)/ε0.
∙(r/r3) = 4πδ(r).
ρ(r) = ρ(r) = 4πε0 A[δ(r) - b exp(-br)/(4πr2)].

We can also use the integral form Gauss' law.
E(r)4πr2 = Qinside0 in SI units.
Qinside = 4πε0 A exp(-br)
ρ(r) = (1/(4πr2))(dQinside /dr) = -ε0 bA exp(-br)/r2, except at r = 0.
As r --> 0 Qinside --> 4πε0 A, there is a point charge at the origin.  It is surrounded by a charge cloud of opposite sign.
Therefore ρ(r) = 4πε0 A[δ(r) -  b exp(-br)/(4πr2)]
(b) As r --> infinite, Qinside --> 0.  The total charge is zero.

Problem 3:

Another set of two thin black parallel plates is inserted between two parallel black surfaces held at temperatures T1 and T2, respectively. 
Calculate the temperatures T' and T'' of these plates in equilibrium. 
(Assume that the area A of the plates is very large, A --> ∞.)

Solution:

bulletConcepts:
Equilibrium, radiation heat transfer, the Stefan-Boltzmann law
bulletReasoning:
The energy radiated by a blackbody per second per unit area is proportional to the fourth power of the absolute temperature, P/A = σT4.
For the plates at temperatures T' and T'' the net power absorbed per unit area must equal the net power emitted per unit area.
bulletDetails of the calculation:
For the plate at temperature T':  T14 T'4 = T'4 T''4.  T''4 = 2T'4 - T14
For the plate at temperature T'':  T'4 T''4 = T''4 T24,  T'4 = -2T14 + 4T'4 T24
T'4 = (2/3)T14 + (1/3)T24,   T''4 = (1/3)T14 + (2/3)T24.

Problem 4:

In bubble chambers, one frequently observes the production of an electron-positron pair by a photon. 
(a)  Show that such a process is impossible unless some other body, for example a nucleus is involved.
(b)  Suppose that the nucleus has mass M and the electron has mass m.  In the rest frame of the electron positron pair, what is the minimum energy the photon must have in order to produce an electron-positron pair?

Solution:

bulletConcepts:
Relativistic collisions, energy and momentum conservation
bulletReasoning:
In relativistic collisions between free particles energy and momentum are always conserved.  Often the physics is best visualized in the center of momentum frame.
bulletDetails of the calculation:
(a)  In the CM frame the total momentum of the electron positron pair is zero and the photon has disappeared.  In that frame the total momentum was also zero before the production of the pair.  The total momentum of a single photon is nonzero, a second object had to be present with momentum equal in magnitude and opposite in direction to the photon momentum..
(b)  The photon has minimum energy if the electron-positron pair is produced with no kinetic energy in the CM frame.
Energy conservation: hf + (M2c4 + p2c2) = 2mc2 + Mc2
Momentum conservation: hf/c = -p.
Combining:  M2c4 + (hf)2 = (2mc2 + Mc2)2 + (hf)2 - 2(2mc2 + Mc2)hf.
hf = ((2mc2 + Mc2)2 - M2c4)/(2(2mc2 + Mc2)) = 2mc2(M + m)/(M + 2m).
If M >> m than the minimum energy of the photon in the CM frame is just slightly less than 2mc2, the kinetic energy of the nucleus contribute the rest.

Problem 5:

 

Two positive and two negative charges are arranged on the corners of a square whose sides have length L.  Each charge has magnitude Q.  The signs of the charges are as indicated.  Point c is at the center of the square.
(a)  What is the electrostatic potential at point c?
(b)  What is the potential energy associated with the charge at point b due to the other three charges?
(c)  How much work must be done to assemble the charges from ∞?  Clearly explain your reasoning, indicate whether the result is + or -, and explain why.

Solution:

bulletConcepts:
Electrostatic potential and potential energy
bulletReasoning:
We use the principle of superposition.
bulletDetails of the calculation:
(a)  V(r)  = k∑qi/|r ri|, with k = 1/(4πε0).
Put the origin of a coordinate system at the center of the square.
Then r = 0 and |r ri] = L/√2 for all charges.  Therefore V(c) = 0.
(b)  V(b) = k[-Q/L + Q/L + Q/(√2 L)] = kQ/(√2 L).
Potential energy of a charge Q at point b:  U = -QV(b) = -kQ2/(√2 L).
(c)  The energy required to assemble a distribution of point charges is
U = k∑all pairs qiqj/|ri rj|.
U = kQ2(1/L 1/(√2 L) 1/L 1/L 1/(√2 L) + 1/L) = kQ2√2/L.
U is negative.