The energy and capacity of a charged parallel plate capacitor are $U$ and $C$ respectively. Now a dielectric slab of dielectric constant $K = 6$ is inserted in it. Then,the new energy and capacity become (Assume the charge on the plates remains constant).

An air capacitor is connected to a battery. The effect of filling the space between the plates with a dielectric is to increase:

If a capacitor with air,a capacitor with a dielectric,and a capacitor with a conducting slab have capacitances $C_1$,$C_2$,and $C_3$ respectively,then:

$A$ capacitor is made of a thin metal strip. Its capacitance is $2 \ \mu F$. If the space between the strips is filled with paper of thickness $0.15 \ mm$,dielectric constant $2.5$,and width $400 \ mm$,what is the length of the strip in meters?

As shown in the figure,a dielectric slab of thickness $d/2$ and dielectric constant $K$ is placed between the plates of a parallel plate capacitor of plate separation $d$. The capacitor is charged to a potential $V$ using a battery. If the dielectric slab is pulled out after disconnecting the battery from the capacitor,the final potential difference across the plates of the capacitor is

The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively $C_0$ and $W_0$. If the air is replaced by glass (dielectric constant $K = 5$) between the plates,the capacity of the plates and the energy stored in it will respectively be