Problem 1

1. The charged ring constitutes a line charge of linear charge density l = Q/2pR. Each (infinitesimally) short segment of length ds contains an amount of charge equal to

dq = l ds = Q/2pR ds . Using the principle of superposition, we can calculate the potential at the point z as the "sum" (really: integral) of the contributions from all these "point charges" dq. If we put the reference point P infinitely far away, then the potential at z due to a single point charge dq will be
dV = 1/4pe_o dq/(z²+R²)^1/2 . Since all point charges dq have the same distance (z²+R²)^1/2 from the point at z=5 cm, all contribute the same amount to the total potential and the sum (integral) is simply over all charges:
V = 1/4pe_{o 1}/(z²+R²)^1/2 Sdq = Q/4pe_o 1/(z²+R²)^1/2 = 1.669^.10⁵ V = 167 kV.
2. Since the "test" charge is negative, it is attracted to the ring and therefore positive work is being done on it as it moves in from infinity. The total amount of work is of course simply given as DW = -q(V(z) - V(°)) = 0.0834J.

The total flux through all six sides of the cube combined is given by Gauss' law as
F_tot = Q_encl/e_o = Q/e_o . Since the charge sits in the center of the cube, each one of the six sides must have the same fraction of that total flux going through it (because of symmetry), which yields 
F_{1 side} = Q/6e_o .

Problem 3

The correct answer is boldfaced - with a short explanation below:

a) The resistivity of copper is 1.7.10^-8 Wm, and that of Aluminum is 2.7^.10^-8 Wm. What diameter does a 1m long Al wire have to have so it has the same resistance as a 1mm diameter 1m long Cu wire? i) 0.63 mm ii) 1.0 mm iii) 1.26 mm iv) 1.59 mm

Since the resistivity of Al is large by a factor 2.7/1.7, the cross sectional area must also be larger by the same factor, and the diameter must be larger by a factor (2.7/1.7)^1/2 .

b) What is the electric potential at the center of a solid conducting sphere of radius
R = 5 cm, with a charge Q = 1.5 nC on it (relative to a point infinitely far away)?

i) 0V ii) 270V iii) 135kV iv) ° (infinitely high)

The potential at the center is the same as the potential at the surface (since it’s a conducting sphere). The latter can be calculated simply using V = Q/4pe_o 1/R since the field outside the sphere is the same as that of a point charge at its center.

c) A parallel-plate capacitor is made of two metal plates with area A=0.2m² at a distance of 1.5mm. How much energy can you store on this capacitor if the maximum voltage may not exceed 1500 V? i) 1.33 mJ ii) 1.5 mJ iii) 1.3^.10^-6 J iv) 1500J

The capacitance can be calculated directly as C = e_o A/d. The stored energy is then 1/2 CV².

d) The electric field 10 m away from a HV power line is 700 V/m. What is the charge per unit length l on that wire? i) 12 nC/m ii) 0.124 µC/m iii) 3.89^.10^-7 C/m
This follows by solving E = l /2pe_o 1/r for l, with E=700V/m and r=10m given.

e) If the power line under d) has a diameter of 5 mm, what would be the electric potential at its surface relative to the point 10 m away? i) 7000 V ii) 58 kV iii) 116 kV
Use V = l /2pe_o ln(r/R) with R = 1/2 diameter = 0.0025m and l from above.