A parallel plate capacitor has a capacitance C_0 when there is air between its plates, which are separated by a distance d . Two dielectric slabs are inserted between the plates, parallel to them. The first slab has a

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Correct answer: C. 2

Initial capacitance is C_0 = ( _0 A)/(d) . When dielectric slabs are inserted, the equivalent capacitance is given by C = ( _0 A)/((t_1)/(K_1) + (t_2)/(K_2) + (t_3)/(K_3)) . The thicknesses of the regions are t_1 = (d)/(2) , t_2 = (d)/(4) , and the remaining air space is t_3 = d - ((d)/(2) + (d)/(4)) = (d)/(4) . Substituting the given values: C = ( _0 A)/((d/2)/(4) + (d/4)/(K) + (d/4)/(1)) C = ( _0 A)/(d ((1)/(8) + (1)/(4K) + (1)/(4))) C = (C_0)/((3)/(8) + (1)/(4K)) Given that C = 2C_0 , we have: 2C_0 = (C_0)/((3)/(8) + (1)/(4K)) (3)/(8) + (1)/(4K) = (1)/(2) (1)/(4K) = (1)/(2) - (3)/(8) = (1)/(8) 4K = 8 K = 2 Answer: 2

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