What is the area of the plates of a capacitor with a capacitance of $3 \ F$? The distance between the two plates is $5 \ mm$.

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:

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 $1\, mm$.

Two circular plates each of radius 'r' are kept parallel to each other distance 'd' apart. The capacitance of the capacitor formed is '$C_1$'. If the radius of each of the plates is increased to $\sqrt{3}$ times the earlier radius and their distance of separation decreased to half the initial value, the capacitance now becomes '$C_2$'. The ratio $C_1 : C_2$ is

Two parallel air capacitors have their plate areas $100 \,cm^2$ and $500 \,cm^2$, respectively. They have the same charge and potential. If the distance between the plates of the first capacitor is $0.5 \,mm$, the distance between the plates of the second capacitor is (in $\,cm$)

$A$ parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge,potential,and capacitance respectively are: