A parallel plate capacitor is made of two circular plates separated by a distance 5mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is `3xx10^4V//m` the charge density of the positive plate will be close to:

Answer

`6`

Explanation

When free space between parallel plates of capacitor
`E=(sigma)/(epsilon_(0))`
When dielectric is intorduced between parallel plates of capacitor `E.=(sigma)/(Kepsilon_(0))`
Electric field inside dielectric
`(sigma)/(K epsilon_(0))=3xx10^(4)`
Where K=dielectric constant of medium =2.2
`epsilon_(0)=` permit avity of free space
`=8.85xx10^(-12)`
`impliessigma=22xx8.85xx10^(-1)xx3xx10^(4)`
`=6xx10^(-7)C//m^(2)`