A parallel capacitor collects a charge of `3muC` when connected to a 1.5 V battery with air as dielectric. On replacing the air with a dielectric material the capacitor collects `9muC` of charge. Find the dielectric constant of the material and also the energy stored in it with the material as dielectric.

Explanation

Given: `V=1.5V,Q_(1)=3xx10^(-6)C,Q_(2)=9xx10^(-6)C,E_(r)=?`
By definition, capacitance
`C=Q/V`
`therefore C=(3xx10^(-6))/(1.5)=2xx10^(-6)F`
Similarly, `C_(med)=epsi_(r)C_(air)`
`therefore epsi_(r)=(6xx10^(-6))/(2xx10^(-6))=3`
energy stored in a capacitor `=(1)/(2)CV_(med)^(2)`
i.e., `E=(1)/(2)xx6xx10^(-6)xx(1.5)^(2)`
i.e., `E=6.75xx10^(-6)J`