If a slab of insulating material $4 \times {10^{ – 3}}\,m$ thick is introduced between the plates of a parallel plate capacitor, the separation between plates has to be increased by $3.5 \times {10^{ – 3}}\,m$ to restore the capacity to original value. The dielectric constant of the material will be

A capacitor with air as the dielectric is charged to a potential of $100\;volts$. If the space between the plates is now filled with a dielectric of dielectric constant $10$, the potential difference between the plates will be……$volts$

Figure given below shows two identical parallel plate capacitors connected to a battery with switch $S$ closed. The switch is now opened and the free space between the plate of capacitors is filled with a dielectric of dielectric constant $3$. What will be the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric

A parallel plate capacitor with air between plates has a capacitance of $8\,\mu F$ what will be capacitance if distance between plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$ ?…..$\mu F$

A parallel plate condenser has a capacitance $50\,\mu F$ in air and $110\,\mu F$ when immersed in an oil. The dielectric constant $'k'$ of the oil is