a parallel plate capacitor has its two plates kept 0. 02 m a part a dielectric slab of thickness 0.01 m is introduced between the plates with its face is parallel to them the distance between the plates is readjusted to make the capacity of the capacitor two third of its original value given that the dielectric constant of the slab is 5 find the new distance between them???
Answers
The new distance between the plates is 0.038 mtrs
Explanation :
We know that capacitance of a parallel plate capacitor without dielectric slab is:
C = ε₀A/d
and capacitance with dielectric slab is :
C' = ε₀A/[d' -t(1 - 1/k)]
since it is given that
C' = 2C/3
=> 2ε₀A/3d = ε₀A/[d' -t(1 - 1/k)]
=> 2/3x0.02 = 1/[d' -0.01(1 - 1/5)]
=> [d' -0.01(1 - 1/5)] = 0.03
=> [d' - 0.01(0.8)] = 0.03
=> d' = 0.038 m
Hence the new distance between the plates is 0.038 mtrs
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