A parallel plate capacitor with air as medium between the plates has a capacitance of $10\,\mu F$. The area of capacitor is divided into two equal halves and filled with two media as shown in the figure having dielectric constant ${k_1} = 2$and ${k_2} = 4$. The capacitance of the system will now be.......$\mu F$

A parallel plate capacitor is made of two circular plates separated by a distance $5\ mm$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3 \times 10^4$ $ Vm^{-1}$ the charge density of the positive plate will be close to

A parallel-plate capacitor of area $A,$ plate separation $d$ and capacitance $C$ is filled with four dielectric materials having dielectric constants $K_1,K_2,K_3$ and $K_4$ as shown in the figure. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $K$ is given by 

A capacitor has some dielectric between its plates and the capacitor is connected to a $\mathrm{D.C.}$ source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.

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

Voltage rating of a parallel plate capacitor is $500\,V$. Its dielectric can withstand a maximum electric field of ${10^6}\,\frac{V}{m}$. The plate area is $10^{-4}\, m^2$ . What is the dielectric constant if the capacitance is $15\, pF$ ? (given ${ \in _0} = 8.86 \times {10^{ - 12}}\,{C^2}\,/N{m^2}$)