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,

$A$ parallel plate capacitor is to be designed with a voltage rating $1\; kV$,using a material of dielectric constant $3$ and dielectric strength about $10^{7}\; V\,m^{-1}$. (Dielectric strength is the maximum electric field a material can tolerate without breakdown,i.e.,without starting to conduct electricity through partial ionisation.) For safety,we should like the field never to exceed,say $10\%$ of the dielectric strength. What minimum area (in $cm^2$) of the plates is required to have a capacitance of $50\; pF$?

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 $5\,mm$?

$A$ parallel plate capacitor is made by stacking $n$ equally spaced plates connected alternately. If the capacitance between any two plates is $C$, then the resultant capacitance is:

$A$ parallel plate capacitor is made of two plates with a diameter of $4 \ cm$. What should be the distance between the two plates so that its capacitance is equal to the capacitance of a sphere with a diameter of $20 \ cm$?