A substance has a dielectric constant 2.0 and its dielectric strength is `20 xx 10^(6)` V/m . It is taken as a dielectric material in a parallel plate capacitor. The minimum area of its each plate such that its capacitance becomes `8.85xx10^(-2) muF ` and it can withstand a potential difference of 2000 V is ......

Explanation

The charge on capacitor
`:. Q =CV = (8.85 xx10^(-8) ) xx(2000)`
`:. Q = 17.7 xx10^(-5) C`
The surface charge density on the plate
`sigma = (Q)/(A) = (17.7xx10^(-5))/(A) C//m^(2)`
The electric field when dielectric inserted
`E = (E_(0))/(K) ` but `E_(0) = (sigma)/(epsilon_(0))`
`:. E = (sigma)/(K epsilon_(0))= (Q)/(K epsilon_(0)A)`
`:. A = (Q)/ (KEepsilon_(0))(17.7xx10^(-5))/(2xx20xx10^(6)xx8.85xx10^(-12))`
`:. A = 0.5 m^(2)`