Five conducting parallel plates having area $A$ and separation between them $d$ are placed as shown in the figure. Plate number $2$ and $4$ are connected by a wire,and between point $A$ and $B$,a cell of emf $E$ is connected. The charge flown through the cell is

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

If the distance between the plates of a parallel plate capacitor of capacity $10 \mu F$ is doubled,then the new capacity will be: (in $\mu F$)

The capacity of a parallel plate capacitor depends on

$A$ large cloud of size $5 \, km \times 5 \, km$ is hovering over the Earth at a height of $1 \, km$. Assuming the cloud and the Earth form a parallel plate capacitor with air as the dielectric,calculate the capacitance of the cloud-Earth system in $\mu F$.

Two identical metal plates are given charges $q_1$ and $q_2$ $(q_2 < q_1)$ respectively. They are brought close together to form a parallel plate capacitor with capacitance $C$. The potential difference $V$ between the plates is