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

Four identical metal plates are located in air at equal distance $d$ from each other. The area of each plate is $S$. If the outer most plates are connected by a conducting wire as shown in the figure, the capacitance between points $A$ and $B$ will be

When the distance between the two plates of a parallel plate capacitor is $6\, cm$,its capacitance is $15\ \mu F$. If the distance between the plates is reduced to $2\, cm$,the capacitance of the parallel plate capacitor becomes ... $\mu F$.

$A$ 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........$V$ $\left( {{\varepsilon _0} = 8.85 \times {{10}^{ - 12}}\,\frac{{{C^2}}}{{N{m^2}}}} \right)$

If the charge on parallel plates is $\pm Q$ and the area is $A$,write the formula for the electric field between the two plates.

Charges of $2Q$ and $-Q$ are placed on two plates of a parallel plate capacitor. If the capacitance of the capacitor is $C$,find the potential difference between the plates.