For a parallel plate capacitor having plates with finite area,the electric field lines bend outward at the edges. This effect is called . . . . . . .

$A$ parallel plate air capacitor is charged to a potential difference of $V$ volts. After disconnecting the charging battery,the distance between the plates of the capacitor is increased using an insulating handle. As a result,the potential difference between the plates:

Four identical plates each of area $a$ are separated by a distance $d$. The connection is shown below. What is the capacitance between $P$ and $Q$?

Why is a $1$ $F$ unit so big in practice?

The intensity of the electric field at a point between the plates of a charged parallel plate capacitor:

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