Question

Q4) (6 Points) A parallel-plate capacitor is constructed by filling the space between two square plates with blocks of four dielectric materials, as shown in the figure below. B) Find the equivalent capacitance, simplify the final expression. A/2 A) Plot the equivalent circuit. A/2 2d

Please answer step-by-step. Thank you!

Answer 1

Alright Dude, If that worked for you... dont forget to give THUMBS UP.(that will work for me!)

If I missed something feel free to leave a comment.
atleast before giving downvote.

Here, C_1 = K_1 epsilon_0 (A/2)/(d/2) = K_1 epsilon_0 A/d = c C_2 = K_2 epsilon_0 (A/2)/(d/2) = (4k_1) epsilon_0 A/d = 4 K_1 epsilon_0 A/d = 4C C_3 = k_3 epsilon_0 (A/2)/(d/2) = (2k_1) epsilon_0 A/d = 2k_1 epsilon_0 A/d = 2C C_4 = K_4 epsilon_0 (A/2)/(d/2) = (6k_1) epsilon_0 A/d = 6K_1 epsilon_0 A/d = 6C Lets take, K_1 epsilon_0 A/d = C So, equivalent of C_1 and C_3: C_13 = C_1 C_3/C_1 + C_3 = (C) (2C)/(C + 2C) = 2C^2/3C = 2/3 C equivalent of C_2 and C_4: C_24 = C_2 C_4/c_2 + c_4 = (4C) (6C)/(4C + 6C) = 24C_2/10 C = 12 C/5 and C_eq = C_13 + C_24 = 2C/3 + 12 C/5 = 10C + 36 C/15 = 46 C/15 So, C_eq = 46/15 k_1 epsilon_0 A/d