If the capacitance of a parallel plate capacitor increases from $10 pF$ to $50 pF$ when a dielectric is introduced, what is the dielectric constant of the material?

The plates of parallel plate capacitor are charged upto $100\;V.$ A $2mm$ thick plate is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by $1.6\;mm.$ The dielectric constant of the plate is

A parallel plate capacitor with air between the plates has a capacitance of $20 \mathrm{~pF}$. The capacitance, if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 3 is:

If dielectric constant and dielectric strength be denoted by $\mathrm{K}$ and $\mathrm{X}$ respectively then a material suitable for use as dielectric in a capacitor must have:

Separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A.$ When a slab of material of dielectric constant $k$ and thickness $t(t < d)$ is introduced between the plates, its capacitance becomes

A parallel plate capacitor with air between the plates has a capacitance of $9\;pF.$ The separation between its plates is $'d'.$ The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant ${k_1} = 3$ and thickness $d/3$ while the other one has dielectric constant ${k_2} = 6$ and thickness $2d/3.$ Capacitance of the capacitor is now

Two parallel plates have equal and opposite charge. When the space between them is evacuated, the electric field between the plates is $4.8 imes 10^5 \mathrm{~V} / \mathrm{m}$. When the space is filled with dielectric, the electric field becomes $1.6 imes 10^5 \mathrm{~V} / \mathrm{m}$. The dielectric constant of the dielectric material is:

A parallel-plate capacitor having plate area $25 \mathrm{~cm}^2$ and separation $1.00 \mathrm{~mm}$ is connected to a battery of $6.0 \mathrm{~V}$. Calculate the charge flown through the battery. How much work has been done by the battery during the process ?

The capacity of parallel capacitor is $10\,\,\mu F.$ When a dielectric medium is introduced between the plates of a parallel plate capacitor is 5. Find the new capacity.

A thin metal plate $P$ is inserted half way 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 is charged by a battery after charging the capacitor, the battery is disconnected and if a dielectric plate is inserted between the plates, then which one of the following statements is not correct