InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A parallel plate capacitor is to be designed with a voltage rating 1 KV using a material of dielectrical 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 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 ? |
|
Answer» Here, `V = 1 kV = 1000 volt , K = in_(r) = 3` , Dielectric strength `= 10^(7) V//m` As electric field at the most should be `10%` of dielectric strength, due to reasons of safety, `:. E = 10% of 10^(7) V//m, A = ?, C = 50 pF = 50xx10^(-12) F` Now, `C = (in_(0) in_(r) A)/(d) A = (Cd)/(in_(0) in_(r)) = (50xx10^(-12)xx10^(-3))/(8.85xx10^(-12)xx3) = 1.9xx10^(-3) m^(2)` |
|