A parallel-plate capacitor is constructed by filling the space between twosquare plates with a block of three dielectric materials, as shown in thefigure below. You may assume that { » d.(a) Find an expression for the capacitance of the device in terms of theplate area A and d, K1, K2, and K3. (Use the following as necessary: K1,K2, K3, E0, A and d.)C =(b) Calculate the capacitance using the values A = 1.00 cm2, d = 2.00mm, K1 = 4.90, K2 = 5.60, and K3 = 2.50.%3DF

Given A = 1.00 cm2d = 2.00 mmk1 = 4.90k2 = 5.60k3 = 2.50

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