Question

5) Find the capacitance per unit length of two parallel wires, each of radius R, separated by some center-to-center distance D. Note that this can be easy or hard depending on whether you're exact or not. The exact approach is to explicitly deal with the fact that the charge distribution on the one wire will affect the charge distribution on the other wire, leading to nonuniform charge distributions on each. The much-easier-but-still-pretty-good approach is to neglect that effect and treat each wire as uniformly charged. I'd invite you to stick to that.

Answer 1

V_A - V_B = integral^d - 2_r E vector dx vector = integral^d - r_r (2 k lambda/x + 2 k lambda/d - x) dx = 2 k lambda ln (d - r/r) + 2 k lambda ln (d - r/r) = 4 k lambda ln (d - r/r) since we know that capacitance c = q/V_A - V_B = q/4 k (q/l) ln (d - r/r) C = pi e_0 l/ln d/r Answer