The two conducting planes illustrated in Figure 6.14 are defined by 0.001 < ρ < 0.120 m, 0 < z < 0.1 m, ϕ = 0.179 and 0.188 rad. The medium surrounding the planes is air. For Region 1, 0.179 < ϕ < 0.188; neglect fringing and find

(a) V(Ï•);

(b) E(ρ);

(c) D(ρ);

(d) ρs on the upper surface of the lower plane;

(e) Q on the upper surface of the lower plane.

(f) Repeat parts (a) through (c) for Region 2 by letting the location of the upper plane be Ï• = .188 ˆ’ 2Ï€, and then find ρs and Q on the lower surface of the lower plane.

(g) Find the total charge on the lower plane and the capacitance between the planes.

p= 12 cm p=1 mm 10 cm Region 2 p = 0.188, V== 20 V Region 1 = 0.179, V= 200 V Gap