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Q.

For charging the capacitance of given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as plates of capacitor. The thickness of dielectric slab is 34d, where d is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance  (C0) is given by 

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a

C'=4C03+K

b

 C'=4KC0K+3

c

C'=(4+K)3C0

d

C'=(3+K)4KC0

answer is A.

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Detailed Solution

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C0=A0d             C1=C1C2C1+C2  or  1C1=1C1+1C2                                     C1=KAε0(3d/4)                    

C2=Aε0(d/4)             1C1=3d/4KAε0+d/4Aε0              C1=4KC03+K

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