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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 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 50 pF? |
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Answer» SOLUTION :Potential rating of a PARALLEL plate capacitor with dielectric = 1 kV = `10^(3)`V For safety the field INTENSITY never exceeds 10 % of the dielectric strength. Hence ELECTRIC field intensity = 10 % of `10^(7)` `= 10(7) xx(10)/(100)` `=10^(6) Vm^(-1)` The minimum necessary distance between two plates `E = (V)/(d) implies d =(V)/(E) = (10^(3))/(10^(6))= 10^(-3)` m Capacitance of parallel plate capacitor `C = (kin_(0)A)/(d)` `:. A = (Cd)/(KE_(0))=(50xx10^(-12)xx10^(-3))/(3xx8.85xx10^(-12))` `:. A = 18.8 xx10^(-4) m^(2)` `:. A = 19 cm^(2)` |
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