$A$ parallel plate capacitor has circular plates of $0.08\,m$ radius and $1.0 \times 10^{-3}\,m$ separation. If a potential difference of $100\,V$ is applied,the charge on the capacitor will be:

The plates of a parallel plate capacitor are pulled apart with a velocity $v$. If at any instant their mutual distance of separation is $d$,then the magnitude of the time rate of change of capacity depends on $d$ as follows:

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

Five identical metal plates, each of area $A$, are arranged parallel to each other at a constant distance $d$ apart. Plates $1$ and $4$ are connected by a wire, and plates $3$ and $5$ are connected by a wire as shown in the figure. Find the equivalent capacitance of the system between points $A$ and $B$.

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

The potential difference between two plates of a parallel plate capacitor is $2V$. As shown in the figure, electrons are placed at point $P$ and $Q$. So,