The capacitance of a parallel plate capacitor with air as medium is `3 muF.` with the introduction of a dielectric medium between the plates, the capacitance becomes `15mu F.` The permittivity of the medium is

Options

A. `5 C^(2)N^(-1) m^(-2)`
B. `15 C^(2)N^(-1) m^(-2)`
C. `0.44 xx 10^(-10) C^(2)N^(-1)m^(-2)`
D. `8.854 xx 10^(-11) C^(2)N^(-1) m^(-2)`

Answer

C

Explanation

Capacitance of a parallel plate capacitor with air is `C = (epsilon_(0)A)/(d)`
Capacitance of a same parallel plate capacitor with the introduction of a dielectric medium is `C. = (K epsilon_(0)A)/(d)`
where K is the dielectric constant of a medium.
or `(C.)/(C ) = K` or `K = (15)/(3) = 5` or `K = (epsilon)/(epsilon_(0))`
or `epsilon = K epsilon_(0) = 5 xx 8.854 xx 10^(-12) = 0.4427 xx 10^(-10)C^(2) N^(-1) m^(-2)`