How will the voltage $(V)$ between the two plates of a parallel plate capacitor depend on the distance $(d)$ between the plates,if the charge on the capacitor remains the same?

If the distance between the plates of a capacitor is halved and the area of the plates is doubled,what will be the new capacitance?

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:

The capacitance of a capacitor made by a thin metal foil is $2\,\mu F$. If the foil is folded with paper of thickness $0.15\,mm$,the dielectric constant of the paper is $2.5$,and the width of the paper is $400\,mm$,then the length of the foil will be.....$m$.

Two protons $A$ and $B$ are placed in the space between the plates of a parallel plate capacitor charged up to $V$ volts (see figure). If the forces on the protons are $F_A$ and $F_B$ respectively,then:

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