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.....$V$
$400$
$800$
$1200$
$1600$
After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates :-
An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air cored parallel capacitor charged to a potential $V$. The two capacitors share charges and the common potential is $V$. The dielectric constant $K$ is
Consider a parallel plate capacitor of $10\,\mu \,F$ (micro-farad) with air filled in the gap between the plates. Now one half of the space between the plates is filled with a dielectric of dielectric constant $4$, as shown in the figure. The capacity of the capacitor changes to.......$\mu \,F$
A parallel plate capacitor has potential $20\,kV$ and capacitance $2\times10^{-4}\,\mu F$. If area of plate is $0.01\,m^2$ and distance between the plates is $2\,mm$ then find dielectric constant of medium
The two metallic plates of radius $r$ are placed at a distance $d$ apart and its capacity is $C$. If a plate of radius $r/2$ and thickness $d$ of dielectric constant $6$ is placed between the plates of the condenser, then its capacity will be