A parallel plate capacitor is to be designed with a voltage rating 1kV, using a material of dielectric constant 3 and dielectric strength about `10^(7)Vm^(-1)`. 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?

Explanation

The maximum electric field applied=10% of dielectric strength `=(10)/(100)xx10^(7)Vm^(-1)=10^(6)Vm^(-1)`
Potential difference across capacitor=1kV=1000V
Capacitance `C=50pF=50xx10^(-12)F`
The maximum charge on the plates `Q=CV=50xx10^(-12)xx1000=5xx10^(-8)C`
If `sigma` is the surface charge density of plates `E=(sigma)/(Kepsi_(0)) implies sigma=Kepsi_(0)E" "therefore Q=sigmaA`
`therefore`Required area `=(Q)/(sigma)=(Q)/(Kepsi_(0)E)=(5xx10^(-8))/(3xx8.85xx10^(-12)xx10^(6))=18.8xx10^(-4)m^(2)=18.8cm^(2)`.