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:

Options

A. `6xx10^(-7)Cm^(-2)`
B. `3xx10^(-7)"Cm"^(-2)`
C. `3xx10^(4)"Cm"^(-2)`
D. `6xx10^(4)"Cm"^(-2)`

Answer

A

Explanation

When there is a free space between parallel plates of capacitor, the electric field, `E=(sigma)/(epsilon_(0))`
When dielectric is introduced between parallel plates of capacitor, `E'=(sigma)/(Kepsilon_(0))`
Electric field inside dielectric, `(sigma)/(Lepsilon_(0))=3xx10^(4)`
where, K = dielectric constant of medium = 2.2
and `" "epsilon_(0)=` permittivity of free space
`=8.85xx10^(-12)C^(2)"Nm"^(2)`
`rArr" "sigma=2.2xx8.85xx10^(-12)xx3xx10^(4)`
`=6.6xx8.85xx10^(-8)=5.84xx10^(-7)`
`=6xx10^(-7)Cm^(-2)`