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 capacity of a parallel plate capacitor is $10\,\mu F$,when the distance between its plates is $8\,cm$. If the distance between the plates is reduced to $4\,cm$,then the capacity of this parallel plate capacitor will be.........$\mu F$.

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:

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

What is the area of the plates of a $3\,F$ parallel plate capacitor,if the separation between the plates is $5\,mm$?

For a parallel plate capacitor having plates with finite area,the electric field lines bend outward at the edges. This effect is called . . . . . . .