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
$K{E_0}$
${E_0}$
$\frac{{{E_0}}}{K}$
None of the above
Condenser $A$ has a capacity of $15\,\mu F$ when it is filled with a medium of dielectric constant $15$. Another condenser $B$ has a capacity of $1\,\mu F$ with air between the plates. Both are charged separately by a battery of $100\;V$. After charging, both are connected in parallel without the battery and the dielectric medium being removed. The common potential now is.....$V$
A combination of parallel plate capacitors is maintained at a certain potential difference When a $3\, mm$ thick slab is introduced between all the plates, in order to maintain the same potential difference, the distance between the plates is increased by $2.4\, mm$. Find the dielectric constant of the slab.
The capacitance of a parallel plate capacitor is $C$ when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant $k$. The capacitor is connected to a cell of $emf$ $E$, and the slab is taken out
The distance between the plates of a parallel plate capacitor is $d$. A metal plate of thickness $d/2$ is placed between the plates. The capacitance would then be
Two parallel metal plates having charges $+Q$ and $- Q$ face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will