$A$ parallel plate air capacitor has capacity $C$ and separation between the plates is $d$. $A$ potential difference $V$ is applied between the plates. The force of attraction between the plates of the parallel plate air capacitor is:

The following arrangement consists of five identical metal plates parallel to each other. The area of each plate is $A$ and the separation between successive plates is $d$. The capacitance between $P$ and $Q$ is

The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If $C$ is its initial capacitance,its final capacitance is equal to

The distance between the plates of a parallel plate capacitor is $4 \ mm$ and the potential difference is $60 \ V$. If the distance between the plates is increased to $12 \ mm$,then:

$A$ parallel plate capacitor was made with two rectangular plates,each with a length of $l=3 \ cm$ and breadth of $b=1 \ cm$. The distance between the plates is $d=3 \ \mu m$. Out of the following,which are the ways to increase the capacitance by a factor of $10$?
$A. l=30 \ cm, b=1 \ cm, d=1 \ \mu m$
$B. l=3 \ cm, b=1 \ cm, d=30 \ \mu m$
$C. l=6 \ cm, b=5 \ cm, d=3 \ \mu m$
$D. l=1 \ cm, b=1 \ cm, d=10 \ \mu m$
$E. l=5 \ cm, b=2 \ cm, d=1 \ \mu m$
Choose the correct answer from the options given below:

Five identical metal plates,each of area $A$,are arranged parallel to each other at a constant distance $d$ apart. Plates $1$ and $4$ are connected by a wire,and plates $3$ and $5$ are connected by a wire as shown in the figure. Find the equivalent capacitance of the system between points $A$ and $B$.