The electric field between the plates of a parallel plate capacitor when connected to a certain battery is $E_0$. If the space between the plates of the capacitor is filled by introducing a material of dielectric constant $K$ without disturbing the battery connections,the field between the plates shall be

$A$ copper slab of thickness $d/2$ is inserted between the plates of a parallel plate capacitor,where $d$ is the distance between its two plates. If the capacitance of the capacitor without the copper slab is $C$ and with the copper slab is $C'$,find the ratio $C'/C$.

$A$ parallel plate capacitor has a capacity of $80 \times 10^{-6} \ F$ when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $K = 20$. The capacitor is connected to a battery of $30 \ V$. The dielectric slab is then removed while the capacitor remains connected to the battery. Calculate the charge that passes through the wire.

$A$ parallel plate capacitor having a separation between the plates $d$,plate area $A$,and material with dielectric constant $K$ has capacitance $C_0$. Now,one-third of the material is replaced by another material with dielectric constant $2K$,such that effectively there are two capacitors: one with area $\frac{1}{3}A$,dielectric constant $2K$,and another with area $\frac{2}{3}A$ and dielectric constant $K$. If the capacitance of this new capacitor is $C$,then $\frac{C}{C_0}$ is:

If a dielectric is inserted into a charged capacitor (battery removed),then the quantity that remains constant is

The plates of a parallel plate capacitor are charged up to $200 \ V$. $A$ dielectric slab of thickness $4 \ mm$ is inserted between its plates. Then,to maintain the same potential difference between the plates of the capacitor,the distance between the plates is increased by $3.2 \ mm$. The dielectric constant of the dielectric slab is