Question

A parallel-plate capacitor is constructed using three dielectric materials, as shown in the figure below. You may assume that e >d ks e/2 (a) Find an expression for the capacitance of the device in terms of the plate area A and d, K, K2 and Click here to answer the question Use A, d, K, K2. K, , and Please enter "epsilon" to write the symbol e.Use a fn) notation to enter an Type your answer (b) Calculate the capacitance using the values A- 1.00 cm?d = 2.50 mm, K, -4.80, K2 - 5.40, and K3 - 2.70 Give your answer in scientific notation rounded to three significant figures. Enter the answer in units. Take the Permittivity of free space to be 8.85 x 10C/N m Click here to answer the question F without the -122 Type your answer

Answer 1

(a) Voltage across the materials K_2 and K_3 will be different that is both dielectrics are connected in services so we can replace both by a single dielectric constant K = K_2 K_3/K_2 + K_3 than the figure will transform to Now voltage across both dielectrics are same that is both are connected in parallel so we can replace both by a single dielectric K_eq = K_1 + K = K_1 + K_2 K_3/K_2 + K_3 K_eq = K_1K_2 + K_1K_3 + K_2K_3/K_2 + K_3 \r\n capacitance of the device is C = K_eq times A epsilon_0/d Now we put value of K_eq. C = (K_1K_2 + K_1K_3 + K_2K_3) A epsilon_0/(K_2 + K_3) d (b) A = 1 cm^2 = 10 ^-4 m^2 d = 2.50 mm = 2.50 times 10^-3 m K_1 = 4.80, K_2 = 5.40, K_3 = 2.70 epsilon_0 = 8.85 times 10^-12 C = (4.80 times 5.40 + 4.80 times 2.70 + 5.40 times 2.70) times