The capacitance of a parallel plate capacitor is $12\,\mu F$. If the distance between the plates is doubled and the area is halved,then the new capacitance will be.........$\mu F$.

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

The unit of absolute permittivity is

$A$ parallel plate capacitor has an electric field of $10^5 \, V/m$ between the plates. If the charge on the capacitor plate is $1 \, \mu C$,the force on each capacitor plate is ...... $N$.

If the plates of a parallel plate capacitor connected to a battery are moved close to each other,then
$A$. the charge stored in it,increases.
$B$. the energy stored in it,decreases.
$C$. its capacitance increases.
$D$. the ratio of charge to its potential remains the same.
$E$. the product of charge and voltage increases.
Choose the most appropriate answer from the options given below:

$A$ 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: