The area A of a parallel plate capacitor is divided into two equal, halves and filled with two media of dielectric constants K_1 and K_2 respectively. The capacitance of the capacitor will be

The area A of a parallel plate capacitor is divided into two equal, ha

1.	The area A of a parallel plate capacitor is divided into two equal, halves and filled with two media of dielectric constants K_1 and K_2 respectively. The capacitance of the capacitor will be
Answer» `(epsi_0 A (K_1 + K_2))/d` `(epsi_0A)/d ((K_1 + K_2)/2)` `(epsi_0A)/d. (K_1K_2)/((K_1 + K_2))` `(epsi_0A)/(2D). (K_1 K_2)/((K_1 +K_2))` Solution :The given arrangement is a COMBINATION of two capacitors in parallel whose CAPACITANCES are `C_1 = (K_1 epsi_0 (A/2))/d andC_2 = (K_2 epsi_0(A/2))/drArrC = C_1 + C_2 = (epsi_0A)/d ((K_1 +K_2)/2)`

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The area A of a parallel plate capacitor is divided into two equal, halves and filled with two media of dielectric constants K_1 and K_2 respectively. The capacitance of the capacitor will be

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