**Problem 1:**

A satellite has a polar orbit about the earth with a radius of 5.49*10^{6}
m above the earth's surface. On one orbit it passes over Greenwich at 0^{o}
longitude and 51.48^{o} latitude. Approximately where will it pass over
exactly one orbit later? The radius of the earth is R_{E} =
6.38*10^{6} m and its mass is 5.98*20^{24} kg.

Solution:

Concepts: Kepler's third law | |

Reasoning: mv ^{2}/r = GMm/r^{2}. v^{2} = GM/r. (2πr/T)^{2}
= GM/r. T ^{2} = (4π^{2}/(GM))r^{3}, Kepler's third law. | |

Details of the calculation: T ^{2} = (4π^{2}/(6.67*10^{-11} * 5.98*10^{24}))(6.38*10^{6}
+ 5.49*10^{6})^{3} = 1.66*10^{8} s^{2}.T = 1.29*10 ^{4} s.The earth rotates through 360 ^{o} in 24 hours and therefore through
53.6^{o} in one period T.One orbit later the satellite will pass over a place with the same latitude and 53.6 ^{o} longitude, west of Greenwich. |

**Problem 2:**

A static charge distribution produces a spherically symmetric, radial
electric field **E** = A exp(-br)/r^{2}(**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:

Concepts: Gauss' law | |

Reasoning:∇∙E = ρ/ε_{0}. For the given field,
the expression is not well defined at r = 0.We can handle the divergence at r = 0 by using ∇∙(r/r^{3})
= 4πδ(r).We can also use the integral form of Gauss' law. | |

Details of the calculation:∇∙E = A exp(-br)∇∙(r/r^{3}) + (r/r^{3})∙∇(A
exp(-br)) = ρ(r)/ε_{0}. ∇∙(r/r^{3}) = 4πδ(r).ρ( r) = ρ(r) =
4πε_{0} A[δ(r) - b exp(-br)/(4πr^{2})].We can also use the integral form Gauss' law. E(r)4πr ^{2 }= Q_{inside}/ε_{0} in
SI units.Q _{inside} = 4πε_{0} A exp(-br)ρ(r) = (1/(4πr ^{2}))(dQ_{inside} /dr) = -ε_{0} bA exp(-br)/r^{2},
except at r = 0.As r --> 0 Q _{inside} --> 4πε_{0} A,
there is a point charge at the origin. It is surrounded by a charge cloud
of opposite sign.Therefore ρ(r) = 4πε _{0} A[δ(r) - b exp(-br)/(4πr^{2})](b) As r --> infinite, Q _{inside} --> 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 T_{1}
and T_{2}, 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:

Concepts: Equilibrium, radiation heat transfer, the Stefan-Boltzmann law | |

Reasoning: The energy radiated by a blackbody per second per unit area is proportional to the fourth power of the absolute temperature, P/A = σT ^{4}.
For the plates at temperatures T' and T'' the net power absorbed per unit area must equal the net power emitted per unit area. | |

Details of the calculation: For the plate at temperature T': T _{1}^{4}
– T'^{4} = T'^{4} – T''^{4}. T''^{4} =
2T'^{4} - T_{1}^{4}For the plate at temperature T'': T' ^{4} – T''^{4} = T''^{4} – T_{2}^{4},
T'^{4} = -2T_{1}^{4} + 4T'^{4} – T_{2}^{4},
T' ^{4} = (2/3)T_{1}^{4} + (1/3)T_{2}^{4},
T''^{4} = (1/3)T_{1}^{4} + (2/3)T_{2}^{4}. |

**
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:

Concepts: Relativistic collisions, energy and momentum conservation | |

Reasoning: In relativistic collisions between free particles energy and momentum are always conserved. Often the physics is best visualized in the center of momentum frame. | |

Details 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 + (M ^{2}c^{4} + p^{2}c^{2})^{½}
= 2mc^{2} + Mc^{2
}Momentum conservation: hf/c = -p.Combining: M ^{2}c^{4} + (hf)^{2} =
(2mc^{2} + Mc^{2})^{2} + (hf)^{2} - 2(2mc^{2}
+ Mc^{2})hf.hf = ((2mc ^{2} + Mc^{2})^{2} - M^{2}c^{4})/(2(2mc^{2} + Mc^{2})) = 2mc^{2}(M + m)/(M + 2m).If M >> m than the minimum energy of the photon in the CM frame is just slightly less than 2mc ^{2}, 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:

Concepts: Electrostatic potential and potential energy | |

Reasoning: We use the principle of superposition. | |

Details of the calculation: (a) V( r) = k∑q_{i}/|r – r_{i}|, with k
= 1/(4πε_{0}).Put the origin of a coordinate system at the center of the square. Then r = 0 and |r – r_{i}] = 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) = -kQ ^{2}/(√2 L).(c) The energy required to assemble a distribution of point charges is U = k∑ _{all pairs }q_{i}q_{j}/|r_{i}
– r_{j}|.U = kQ ^{2}(1/L – 1/(√2 L) – 1/L – 1/L – 1/(√2 L) + 1/L) = –kQ^{2}√2/L.U is negative. |