Two identical metal plates are given charges $q_1$ and $q_2$ $(q_2 < q_1)$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance $C$,the potential difference $V$ between the plates is

$A$ parallel plate capacitor with area $200\,cm^2$ and separation between the plates $1.5\,cm$,is connected across a battery of $emf$ $V$. If the force of attraction between the plates is $25\times10^{-6}\,N$,the value of $V$ is approximately........$V$ $\left( {{\varepsilon _0} = 8.85 \times {{10}^{ - 12}}\,\frac{{{C^2}}}{{N{m^2}}}} \right)$

The plates of a parallel plate capacitor are pulled apart with a velocity $v$. If at any instant their mutual distance of separation is $d$,then the magnitude of the time rate of change of capacity depends on $d$ as follows:

In a parallel plate capacitor,the capacity increases if

$A$ parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved further apart by pulling them by means of insulating handles,then

The capacity of a parallel plate capacitor is $10\,\mu F$,when the distance between its plates is $8\,cm$. If the distance between the plates is reduced to $4\,cm$,then the capacity of this parallel plate capacitor will be.........$\mu F$.