If the distance between the plates of a parallel plate capacitor of capacity $10 \mu F$ is doubled,then the new capacity will be: (in $\mu F$)

For a parallel plate capacitor having plates with finite area,the electric field lines bend outward at the edges. This effect is called . . . . . . .

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:

Charges of $2Q$ and $-Q$ are placed on two plates of a parallel plate capacitor. If the capacitance of the capacitor is $C$,the potential difference between the plates is:

Six metallic plates,each with a surface area $A$ on one side,are placed at a distance $d$ from each other. The alternate plates are connected to points $P$ and $Q$ as shown in the figure. The capacitance of the system is:

$A$ parallel plate capacitor has circular plates of $10\, cm$ radius separated by an air-gap of $1\, mm$. It is charged by connecting the plates to a $100\, V$ battery. The change in energy stored in the capacitor when the plates are moved to a distance of $1\, cm$ while remaining connected to the battery is: