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Two dielectric slabs of dielectric constants K_1 and K_2 are filled in between the two plates, each of area A, of the parallel plate capacitor as shown in the figure. Find the net capacitance of the capacitor. |
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Answer» SOLUTION :The arrangement is EQUIVALENT to two capacitors JOINED in parallelwhere area of plates of either capacitor is `A/2`. THUS, `C_1 =(K_1 epsi_0 (A/2))/d = (K_1 epsi_0A)/(2d)` and `C_2 =(K_2epsi_0(A/2))/(d) =(K_2 epsi_0A)/(2d)` `:.` NET capacitance of the capacitor `C = C_1 + C_2= (K_1 epsi_0 A)/(2d) + (K_2 epsi_0A)/(2d) = (epsi_0A)/d ((K_1 + K_2)/2)`. |
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