The charge on a parallel plate capacitor of capacity $C$ is $Q$. The electric field intensity between its two plates,which are separated by a distance $t$,is:

The capacity of a parallel plate capacitor is $15\,\mu F$,when the distance between its plates is $6\,cm$. If the distance between the plates is reduced to $2\,cm$,then the capacity of this parallel plate capacitor will be......$\mu F$.

The lower plate of a parallel plate capacitor is supported on a rigid rod. The upper plate is suspended from one end of a balance. The two plates are joined together by a thin wire and subsequently disconnected. The balance is then counterpoised. Now a voltage $V = 5000 \, V$ is applied between the plates. The distance between the plates is $d = 5 \, mm$ and the area of each plate is $A = 100 \, cm^2$. Find the additional mass $m$ (in grams) placed to maintain balance. [All elements other than plates are massless and nonconducting. Take $\epsilon_0 = 8.85 \times 10^{-12} \, F/m$ and $g = 9.8 \, m/s^2$].

Two identical thin metal plates have charges $q_{1}$ and $q_{2}$ respectively,such that $q_{1} > q_{2}$. The plates are brought close to each other to form a parallel plate capacitor of capacitance $C$. The potential difference between them is:

$A$ parallel plate capacitor is charged by a battery and then the battery is disconnected. If the distance between the plates is increased using an insulated handle,then:

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