Question

(11) The following figures show two parallel-plate capacitors with both charge Qo Qo E2 D? (a) As shown in picture (a), the length of the capacitor is a, and width is b. The top part of the gap is filled with material of dielectric constant ε1, and bottom part with material of dielectric constant ε2. The space between the two metal plate is d - di + d2. Calculate the E field and D field in both materials. (b) As shown in picture (b), The length of the capacitor is a, and width is b- bı + b2. The left part of the gap is filled with material of dielectric constant e1, and right part with material of dielectric constant E2. The space between the two metal plate is d. Calculate the E field and D field in both materials. (c) According to your calculation in above questions, show following formulas for series capacitor and parallel capacitor is true for series capacitor C-C1 + C2 for parallel capacitor [hints: considering boundary conditions, does E = E2 or D1 D, in both situations? [Total: 12 marks]

Answer 1

For (a) E vector = Q/epsilon_0 A, D vector = epsilon E vector In (a), top part of parallel - plate capacitor dielectric constant = epsilon_0 Area of this top part = length times width A_1 = a times b A_1 = (a - d_1) b Therefore, E_1 vector = Q_0/epsilon_(a - d_1) b and D_1 vector = epsilon_1 E_1 vector = > D_1 vector = Q_0/(a - d_1) b In (a), bottom part of parallel - plate capacitor dielectric constant = epsilon_2 Area of this bottom part = length times width A_2 = a times b A_2 = (a - d_2) b Therefore, E_2 vector = Q_0/epsilon_2 (a - d_2) b And D_2 vector = epsilon_2 E_2 vector = > D_2 vector = Q_0/(a - d_2) b \r\n For (b) In (b) left part of parallel plate capacitor. Dielectric constant = epsilon_1 length = a and width for this left portion is b_1 So Area = a times b_1