Question
Question: A parallel plate capacitor of capacitance \(51pF\) is too designed with voltage rating \(1kV\) using...
A parallel plate capacitor of capacitance 51pF is too designed with voltage rating 1kV using a material of dielectric constant 200 and dielectric strength about 107Vm−1 . If the electric field is never to exceed 10% of dielectric strength what should be the minimum area of plates?
Solution
In order to solve this question we need to understand the definition of capacitor. Capacitor is an electrical device that is used to store electric energy in form of charges stored on its plates and later used in some circuits as a source of energy. There are three types of capacitors used, one is parallel plate capacitor, second is cylindrical plate capacitor and the last one is spherical plate capacitor.
Complete step by step answer:
According to the question we have, given capacitance C=51pF dielectric constant k=200 dielectric strength E=107Vm−1 and VoltageV=1kV Also ε0=8.854×10−12C2N−1m−2.
Since value of electric field permitted to use is 10% so electric field across plates is E′=10010×107Vm−1
E′=106Vm−1
Now let the plate separation be “d” so by using relation E=dV we get
d=EV=106Vm−1103V
⇒d=10−3m
Let plate area be “A” then by using C=dkε0A
Putting values we get
A=kε0Cd
⇒A=(200×8.854×10−12)(10−12×10−3)
∴A=0.056×10−5m2
So the minimum plate is area=0.056μm2.
Note: It should be remembered that dielectric strength is the electric field inside dielectric whereas dielectric is defined as an insulator which gets polarized under the influence of external electric field and thereby generating displacement vector. Also dielectric constant is a measure of resistance shown by insulators in response to their behavior in an external electric field. Capacitance increases if we place a dielectric material between the plates of the capacitor and 1μm=10−6m.