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A parallel plate capacitor is to be designed with a voltage rating 1 KV using a material of dielectric constant 3 and dielectric strength about 10^(7) Vm^(-1). [Dielectric strength is the maximum electric field a material can tolerate without break down, i.e, without starting to conduct electrically through partial ionization. For safety, we should like the field never to exceed say 10% of the dielectric strength]. What minimum area of the plates is required to have a capacitance of 50 pF ?

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Solution :
Potential rating of a parallel plate capacitor , V=1kv =1000 V <br> dielectric constant of a meterial `in_(r)=3` <br> dielectric strength `=10^(7)V//m` <br> For safety, the field intensity never exceeds 10 % of the dielectric strength. Hence, electric field intensity, `E=10 % ` of `10^(7) =10^(6) V//m` <br> Capactiance of the parallel plate capacitor , C=50 pF =`50xx10^(-12)F` <br> distance between the plates in given by , <br> `d=V/E` <br> `=1000/(10^(6))=10^(-3) m` <br> Capacitance is given by the relation, <br> `C=(in_(o)in_(r)A)/d` <br> where, A= area of each plate <br> `in_(0)`=permitivity of free space =`8.85xx10^(-12) N^(-1)C^(2)m^(-2)` <br> `:. A=(Cd)/(in_(0)in_(r))` <br> `=(50xx10^(-12)xx10^(-3))/(8.85xx10^(-17)xx3) ~~19cm^(2)` <br> Hence, the area of each plate is about `19 cm^(2)`.

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