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 area of each plate of a parallel plate capacitor is $20 \, cm^2$ and the separation between the plates is $2 \, mm$. If the dielectric strength of air is $3 \times 10^6 \, V/m$,the maximum possible value of the emf of the battery that can be connected across the plates of this capacitor and the corresponding charge on the plates are:

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 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,

One plate of a parallel plate capacitor is connected to a spring as shown in the figure. The area of each plate of the capacitor is $A$ and the distance between the plates is $d$,when the battery is not connected and the spring is unstretched. After connecting the battery,in the steady state the distance between the plates is $0.75 d$,then the force constant of the spring is

Five identical capacitor plates are arranged as shown in the figure. They form capacitors each of $2 \mu F$. The plates are connected to a source of $emf$ $10 \ V$. The charge on plate $C$ is......$\mu C$.