A parallel plate capacitor with air between the plates has a capacitance of 8 pF.(1 pF = 10^(-12)F) What will be the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 6 ?
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Capacitance between the parallel plates of the capacitor, C=8 pF <br> initially, distance betwene the parallel plates was d and it was filled with air. Dielectric constant of air, k=1 <br> capacitance, C is given by the formula. <br> `C=(kin_(0)A)/d` <br> `=(in_(0)A)/d....(i)` <br> Where, A=Area of each plate <br> `in_(0)`=permitivity of free space <br> IF distance between the plates is reduced to half , then new distance `d=(d/2)` <br> dielectric constant of the substance filled in between the paltes `k=6` <br> Hence, capacitance of the capacitor becomes <br> `C"=(k in_(0)A)/(d')=(6in_(0)A)/(d/2).....(ii)` <br> Takin ratios equation (i) and (ii) , we obtain <br> `C'=2xx6C` <br> `=12C` <br> `=12xx8=96 pF` <br> Therefore, the capacitance between the paltes is 96 pF