When a test charge placed in the middle of a parallel plate capacitor experiences a force $F$,what will be the force experienced by the same test charge if one plate is removed?

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?

When the distance between the two plates of a parallel plate capacitor is $6\, cm$,its capacitance is $15\ \mu F$. If the distance between the plates is reduced to $2\, cm$,the capacitance of the parallel plate capacitor becomes ... $\mu F$.

In an isolated 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

The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If $C$ is its initial capacitance,its final capacitance is equal to

$A$ parallel plate capacitor is to be designed,using a dielectric of dielectric constant $5$,so as to have a dielectric strength of $10^9 \; Vm^{-1}$. If the voltage rating of the capacitor is $12 \; kV$,the minimum area of each plate required to have a capacitance of $80 \; pF$ is