$A$ metal slab of thickness $b = d/2$ is inserted between the plates of a parallel plate capacitor of plate separation $d$. What is the ratio of the new capacitance to the original capacitance?

Two parallel plate capacitors with different plate separations but the same capacitance are connected in series to a battery. Both capacitors are filled with air. The quantity that is $NOT$ the same for both the capacitors when they are fully charged is:

Five identical plates each of area $A$ are joined as shown in the figure. The distance between the plates is $d$. The plates are connected to a potential difference of $V \text{ volts}$. The charge on plates $1$ and $4$ will be

$A$ capacitor is made of a flat plate of area $A$ and a second plate of stair-like structure as shown in the figure. The area of each stair is $\frac{A}{3}$ and the height is $d$. The capacitance of the arrangement is

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to $\frac{1}{2} Q E$,where $Q$ is the charge on the capacitor,and $E$ is the magnitude of the electric field between the plates. Explain the origin of the factor $\frac{1}{2}$.

$A$ parallel plate capacitor with area $200\,cm^2$ and separation between the plates $1.5\,cm$,is connected across a battery of $emf$ $V$. If the force of attraction between the plates is $25\times10^{-6}\,N$,the value of $V$ is approximately........$V$ $\left( {{\varepsilon _0} = 8.85 \times {{10}^{ - 12}}\,\frac{{{C^2}}}{{N{m^2}}}} \right)$