$A$ copper slab of thickness $d/2$ is inserted between the plates of a parallel plate capacitor,where $d$ is the distance between its two plates. If the capacitance of the capacitor without the copper slab is $C$ and with the copper slab is $C'$,find the ratio $C'/C$.

When a dielectric substance is placed between the two plates of a capacitor,what happens to its capacitance,potential difference,and potential energy,respectively?

$A$ parallel plate capacitor whose capacitance $C$ is $14 \, pF$ is charged by a battery to a potential difference $V = 12 \, V$ between its plates. The charging battery is now disconnected and a porcelain plate with dielectric constant $k = 7$ is inserted between the plates. The plate would oscillate back and forth between the plates with a constant mechanical energy of $.......... pJ$. (Assume no friction)

$A$ thin metal plate $P$ is inserted halfway between the plates of a parallel plate capacitor of capacitance $C$ in such a way that it is parallel to the two plates. The capacitance now becomes

$A$ parallel plate capacitor has a parallel sheet of copper inserted between and parallel to the two plates,without touching the plates. The capacity of the capacitor after the introduction of the copper sheet is:

$A$ capacitor has air as dielectric medium and two conducting plates of area $12 \,cm^2$ and they are $0.6 \,cm$ apart. When a slab of dielectric having area $12 \,cm^2$ and $0.6 \,cm$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \,cm$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $\epsilon_0 = 8.834 \times 10^{-12} \,F/m$)