Question

(C) Calculate the total capacitance of the structure that consists of a dielectric slab (relative permittivity is ε) inserted between two parallel plates that are separate by a distance D. The thickness of the dielectric slab is d (obviously d 〈 D). The widths of the two gaps between the dielectric and two electrodes are (D - d)/3 and (2D - 2d)/3. Neglect fringing field effect and assume that the areas are all equal to A. AND when it is connected to a voltage source U, what is the total electrostatic energy stored in the system?

Answer 1

q = cv c = q/v v = v_a + v_m cross sectional Area = A where v_m is the voltage the slab v = E_0 (D - d) + E middot d (Therefore E = E_0/k). = E_0 (D - d) + E_0/k middot d = E_0 (D - d + d/k) (Therefore D - d = D - d/3 + 2D - 2d/3 = D/3 - d/3 + 2D/3 - 2d/3) = D[1/3 + 2/3] - d [1/3 + 2/3] = [D - d] v = sigma/epsilon_0 [D - d + d/k] c = q/v = q/sigma/epsilon_0 (D - d + d/k) = q/q/A epsilon_0 (D - d + d/k) = 1/1/A epsilon_0 [D - d + d/k] = c E = 1/2 cv^2 = 1/2 [1/1/A epsilon_0 (D - d + d/k)] u^2