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A parallel plate capacitor is to be designed with a voltage rating 1kV, using a maeterial of dielectric constant 3 and dielectric strength about 10^(7) Vm^(-1), (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e, without starting to conduct electricity through partial ionisation.) 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 50pF? |
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Answer» Solution :`V=10^(3)V, epsi_(r)=3, ""E_(de)=10^(7)VM^(-1),""E=(10)/(100) xx 10^(7) =10^(6)Vm^(-1)` `A=? C=50pF =50 xx 10^(-12)F` `E=V/d, therefore d=V/E=(10^(3))/(10^(6))=10^(-3)m` `C=(epsi_(0)epsi_(r)A)/(d)` `A=(C.d)/(epsi_(0) epsi_(r))= (50 xx 10^(-12) xx 10^(-3))/(8.85 xx 10^(-12) xx 3)=1.9 xx 10^(-3) m^(2)=19 xx 10^(-4) m^(2)=19CM^(2)` |
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