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:

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?

In a parallel plate capacitor with air between the plates,each plate has an area of $6 \times 10^{-3} \, m^{2}$ and the distance between the plates is $3 \, mm$. Calculate the capacitance of the capacitor. If this capacitor is connected to a $100 \, V$ supply,what is the charge on each plate of the capacitor?

The distance between the circular plates of a parallel plate capacitor $40\,mm$ in diameter,in order to have the same capacity as a sphere of radius $1\,m$,is....$mm$.

In an insulated parallel-plate capacitor of capacitance $C$,the four surfaces have charges $Q_1, Q_2, Q_3$ and $Q_4$ as shown. The potential difference between the plates is

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 $\mu F$)