$A$ metal slab of thickness $b = d/2$ is inserted between the plates of a parallel plate capacitor of plate separation $d$. What is the ratio of the new capacitance to the original capacitance?

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

In a parallel plate capacitor,if $10^{12}$ electrons pass from one plate to another,a potential difference of $10 \,V$ is developed across the plates. The capacitance of the capacitor is

$A$ 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:

If the charge on each plate of a parallel plate capacitor is $Q$ and the magnitude of the electric field between the plates is $E$,then the force on each plate of the parallel plate capacitor will be:

$A$ parallel plate capacitor is connected to a battery,which maintains a constant potential difference. If the plates of the capacitor are moved further apart,the electric field intensity...