A parallel$-$plate capacitor with plate area A $=2\mathrm{.}0$m$2$ and plate separation d$=$

$3\mathrm{.}0$mm is connected to a $45$V battery.

$($a$)$ Determine the charge on the capacitor, the electric field, the capacitance,

and the energy stored in the capacitor.

$($b$)$ With the capacitor still connected to the battery, a slab of plastic with

dielectric strength K $=3\mathrm{.}2$ is placed between the plates of the capacitor, so

that the gap is completely filled with the dielectric. What are the new values of

charge, electric field, capacitance, and the energy stored in the capacitor?

ANS: $5\mathrm{.}9109$F$,2\mathrm{.}7107$C$,15000$V$/$m$,6\mathrm{.}0106$J

$1\mathrm{.}9108$F$,8\mathrm{.}5107$C$,15000$V$/$m$,1\mathrm{.}9105$J