InterviewSolution
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InterviewSolution
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A dielectric slab of thickness 't' is kept between the plates of a parallel plate capacitor separated by a distance 'd' (t lt d) . Derive the expression for the capacity of the capacitor . |
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Answer» Solution :In the absence of dielectric slab , the capacitance of a PARALLEL PLATE capacitor is given by `C = (in_(0) A)/(d)` When a dielectric slab of THICKNESS `t(t lt d)` is introduced between the PLATES without touching the plates , the electric field in air `E_(0) = (sigma)/(in_(0))` but on account of polarisation of dielectric the electric field inside the dielectric changes to `E = (E_0)/(K)` . If potential difference between the plates of capacitor be V. now , then clearly `V = E_0 (d- t) + E t = E_(0) (d- t) + Et = E_(0) (d - t) + (E_(0))/(K) . t = E_(0) (d - t + (t)/(K)) = (sigma)/(in_(0)) . (d - t + (t)/(K)) = (Q)/(in_(0) A) (d - t + (t)/(K))`  `therefore` New capacitance `C. = (Q)/(V.) = (Q)/((Q)/(in_(0) A (d - t + (t)/(K))) = (in_(0) A)/(d - t ((K-1)/(K))) = (in_(0) A)/(d[1 - (t)/(d) ((K-1)/(K))]) = (C)/(1 - (t)/(d) ((K-1)/(K)))` |
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