Question

pression for the capacitance per unit area of a parallel plate capacitor, of tb d) and plate separation d when two dielectric materials of width a and b (a electric constants , and ?2, respectively, are placed inside the capacitor expression should be as a function of The e1, 82, a, and b. (See Fig. Q6). 18 marks 8 marks

Answer 1

This Capacitance acts as two capacitors in series as shown below rightarrow For region with epsilon = epsilon_1 separation = a, capacitor c_1 = epsilon_1 A_p epsilon_0/a rightarrow For region with epsilon = epsilon_2 & separation = d, capacitor c_2 = epsilon_2 A_p epsilon_0/b For capacitance in series, 1/c_q = 1/c_1 + 1/c_2 = a/epsilon_1 A_p epsilon_0 + b/epsilon_2 A_p epsilon_0 1/c_q = 1/a_p epsilon_0 [a/A_p epsilon_0 [a/epsilon_1 + b/epsilon_2] = a epsilon_2 + b epsilon_1/A-p epsilon_0 epsilon_1 epsilon_2 c_q = epsilon_0 A_p epsilon_1 epsilon_2/a epsilon_2 + b epsilon_1 capacitance per unit area implies c_q/A_p = epsilon_0 epsilon_1 epsilon_2/a epsilon_2 + b epsilon_1

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