hwk 8-1. Given the capacitor as described, the electric field is constant everywhere inside the plates and directed left to right. It's value is E = Q/(ıR² e₀) .

Since DV= -E. DL = - E DL cos u, where u is the angle between E and DL DV_DC = DV_AD = 0, (cuz u = 90°), anD
a-1) DV_ABC = DV_AB + DV_BC = - E s₁ cos(180°) + 0 = E s₁.

a-2) DV_AC = -E DL cos u = -E (-s1) = E s₁ (see figure at right )

a-3) DV_ADBC = DV_AD + DV_DB + DV_BC
= 0 + (-E)(-s₁) + 0 = E s₁

b) DVAC =

Hwk8-2: Have DV_AB = 1.5 V

Which end has higher potential? B
Magnitude and direction of E inside the resistor? E = 1.5V / 10-2 m = 1500 V/m toward A.