A parallel plate condenser with a dielectric of dielectric constant $K$ between the plates has a capacity $C$ and is charged to a potential $V\ volt$. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is
Zero
$\frac{1}{2}\left( {K - 1} \right)C{V^2}$
$\frac{{C{V^2}\left( {K - 1} \right)}}{K}\;\;\;\;\;$
$\;\left( {K - 1} \right)C{V^2}$
Two dielectric slabs of constant ${K_1}$ and ${K_2}$ have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor
Two condensers of capacities $2C$ and $C$ are joined in parallel and charged upto potential $V$. The battery is removed and the condenser of capacity $C$ is filled completely with a medium of dielectric constant $K$. The $p.d.$ across the capacitors will now be
Polar molecules are the molecules:
A parallel plate condenser has a capacitance $50\,\mu F$ in air and $110\,\mu F$ when immersed in an oil. The dielectric constant $'k'$ of the oil is
Following operations can be performed on a capacitor : $X$ - connect the capacitor to a battery of $emf$ $E.$ $Y$ - disconnect the battery $Z$ - reconnect the battery with polarity reversed. $W$ - insert a dielectric slab in the capacitor