As shown in the figure,two parallel plate capacitors having equal plate area of $200 \, cm^2$ are joined in such a way that $a \neq b$. The equivalent capacitance of the combination is $x \varepsilon_0 \, F$. The value of $x$ is $..........$.

The plates of a parallel plate capacitor of capacity $C_1$ are moved closer together until they are half their original separation. The new capacitance $C_2$ is

$A$ 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:

$A$ parallel plate capacitor is charged by a battery and then the battery is disconnected. If the distance between the plates is increased using an insulated handle,then:

$A$ parallel plate capacitor consisting of two circular plates of radius $10 \ cm$ is being charged by a constant current of $0.15 \ A$. If the rate of change of potential difference between the plates is $7 \times 10^8 \ V/s$,then the integer value of the distance between the parallel plates is $—$ (Take $\epsilon_0 = 9 \times 10^{-12} \ F/m, \pi = 22/7$) . . . . . . $\mu m$.

$A$ 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: