Five identical capacitor plates,each of area $A$,are arranged such that adjacent plates are at a distance $d$ apart. The plates are connected to a source of $emf$ $V$ as shown in the figure. Then the charges on plates $1$ and $4$ are,respectively:

What is the area of the plates of a $2\,\mu F$ parallel plate capacitor,given that the separation between the plates is $0.5\, cm$?

The electric field between the plates of a parallel plate capacitor is $E$. If the charge on the plates is $Q$, then the force on each plate is:

$A$ parallel plate air capacitor is charged to a potential difference of $V \text{ volts}$. After disconnecting the charging battery,the distance between the plates of the capacitor is increased using an insulating handle. As a result,the potential difference between the plates

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 ... $\mu F$.

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 $g = 9.8 \, m/s^2$].