Find the distance $d$ (in $mm$) between the plates of a parallel plate capacitor with a plate diameter of $40\, mm$,such that its capacitance is equal to the capacitance of a metallic sphere of radius $1\, mm$.

$A$ parallel plate capacitor is to be designed,using a dielectric of dielectric constant $5$,so as to have a dielectric strength of $10^9 \; Vm^{-1}$. If the voltage rating of the capacitor is $12 \; kV$,the minimum area of each plate required to have a capacitance of $80 \; pF$ is

One plate of a parallel plate capacitor is connected to a spring as shown in the figure. The area of each plate of the capacitor is $A$ and the distance between the plates is $d$,when the battery is not connected and the spring is unstretched. After connecting the battery,in the steady state the distance between the plates is $0.75 d$,then the force constant of the spring 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:

Four metallic plates,each with a surface area of one side $A$,are placed at a distance $d$ from each other. The plates are connected as shown in the circuit diagram. Then the capacitance of the system between $a$ and $b$ is

$A$ capacitor of capacitance $C$ with a distance $d$ between its plates is connected to a battery of $V$ volts. What is the force acting between the two plates?