Calculate the capacitance of two plates of equal area of $1 \ m^2$ separated by a distance of $1 \ mm$.

What is a parallel plate capacitor? Obtain the formula for the capacitance of such a parallel plate capacitor and state the factors on which the value of capacitance depends.

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

Two identical thin metal plates have charges $q_{1}$ and $q_{2}$ respectively,such that $q_{1} > q_{2}$. The plates are brought close to each other to form a parallel plate capacitor of capacitance $C$. The potential difference between them is:

$A$ 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 charges on the four surfaces of two isolated plates with capacitance $C$ are $Q_1$,$Q_2$,$Q_3$,and $Q_4$ as shown in the figure. The potential difference between the plates is: