A parallel plate capacitor with plate area A and plate separation d=2 ~m has a capacitance of 4 ~F . The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant K =3 (as shown in figure) will be:

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Correct answer: C. 6 ~F

We have, C_i= A _0 d =4 F C_f= A _0 d-t+ t k = A _0 d- d 2 + d 2 3 = A _0 d (1- 1 2 + 1 6 ) = 4 F 2 3 =6 F

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