One plate of a parallel plate capacitor is smaller than the other. What can be said about the charge on the smaller plate?

The force acting on a charged particle placed between the plates of a charged parallel plate capacitor is $F$. If one plate of the capacitor is removed,then the force acting on the same particle will become:

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to $\frac{1}{2} Q E$,where $Q$ is the charge on the capacitor,and $E$ is the magnitude of the electric field between the plates. Explain the origin of the factor $\frac{1}{2}$.

In a parallel plate capacitor,the capacity increases if

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

The insulated plates of a charged parallel plate capacitor (with small separation between the plates) are approaching each other due to electrostatic attraction. Assuming no other force to be operative and no radiation taking place, which of the following graphs approximately shows the variation with time $(t)$ of the potential difference $(V)$ between the plates?