A parallel plate capacitor having cross-sectional area A and separation d has air in between the plates. Now an insulating slab of same area but thickness d/2 is inserted between the plates as shown in figure having dielectric constant K(= 4). The ratio of new capacitance to its original capacitance will be,
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a
4 : 1
b
2 : 1
c
8 : 5
d
6 : 5
answer is C.
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Detailed Solution
C0=ε0AdAfter inserting dielectricC=ε0A(d−t)+tk=ε0Ad2+d8=8ε0A5d=85C0 So, CC0=85