Effect of Dielectrics on Capacitance

The two plates of a parallel plate capacitor are $4 \mathrm{mm}$ apart. A slab of dielectric constant $3$ and thickness $3 \mathrm{mm}$ is introduced between the plates with its faces parallel to them. The distance between the plates is so adjusted that the capacitance of the capacitor becomes ${\left(\frac{2}{3}\right)}^{rd}$ of its original value. What is the new distance between the plates?

When a conducting slab is kept between two plates of parallel plate capacitor then capacitance.

A conducting slab has kept between two plates of parallel plate capacitor then electric field inside the slab is

A parallel plate capacitor of capacitance $90 \mathrm{pF}$ is connected to a battery of emf $20 \mathrm{V}$. If a dielectric material of dielectric constant $K=\frac{5}{3}$ is inserted between the plates, the magnitude of the induced charge will be:

An electron with a kinetic energy of $100\mathrm{ }\mathrm{e}\mathrm{V}$ enters the space between the plates of the plane capacitor made of two dense metal grids at an angle of $30°$ with the plates of capacitor and leaves this space at an angle of $45°$ with the plates. What is the potential difference of the capacitor?

Two conducting square plates with length $l$ are arranged parallel to each other at a distance $d(\ll 1)$. The space between the plates is filled with a dielectric of relative permittivity ${\epsilon }_{\mathrm{r}}=4$ up to a length $x=l/3$. The upper plate is given a charge $Q$ and the lower plate is given a charge $-Q$ as shown in the figure. If $Q=6 \mu \mathrm{C}$ then find how much charge (in $\mu \mathrm{C}$) is present on the portion of upper plate which is in contact with the dielectric?

A parallel plate capacitor is connected to a battery which builds up an electric field of $60 \mathrm{V} {\mathrm{cm}}^{-1}$ between the plates as shown in figure-I. Now two initially neutral plates are positioned and connected as shown in figure-II. The plates are at equal distances from each other. Find the strength of electric field (in $\mathrm{kV} {\mathrm{m}}^{-1}$) between the plates $B$ and $D.$

A parallel plate capacitor of capacitance $C$ has spacing $D$ between two plates having area $A.$ The region between the plates is filled with $N$ dielectric layers, parallel to its plates, each with thickness $\delta =\frac{D}{N}.$ The dielectric constant of the $m^{th}$ layer is $K_m=K\left(1+\frac{m}{N}\right).$ For a very large $N\left(>{10}^3\right),$ the capacitance $C$ is $\alpha \left(\frac{k{\epsilon }_0 A}{ D\ln 2}\right).$ The value of $\alpha$ will be - [${\epsilon }_0$ is the permittivity of free space]

Capacity of a parallel capacitor with dielectric constant $5$ is $40 \mathrm{\mu F}$. Capacity of the same capacitor in $\mathrm{\mu F}$ when dielectric material is removed is

The capacitance of a parallel plate air capacitor is $C_0$. If a dielectric of dielectric constant $K$, and thickness equal to one fourth the separation is placed between the plates, then capacity becomes $C$. Then, the ratio $\frac{C}{C_0}$ is,

A parallel plate capacitor with plate area $1 {\mathrm{m}}^2$ and plate separation $1 \mathrm{mm}$ is charged at the rate of $25 \mathrm{V} {\mathrm{s}}^{-1}$. The dielectric between the plates has a dielectric constant $k$. If the displacement current through the capacitor is $2.21 \mathrm{\mu }$ A then the value of $k^{\text{'}}$ is nearly

A parallel plate capacitor has a capacity $80\times {10}^{-6} \mathrm{F}$ when air is present between its plates. The space between the plates is filled with a dielectric slab of dielectric constant $20$. The capacitor is now connected to a battery of $30 \mathrm{V}$ by wires. The dielectric slab is then removed. Then the charge passing through the wire is

In a parallel plate with air capacitor of capacitance $8 \mu F,$ if the lower half of the air space is filled with a material of dielectric constant $3,$ its capacitance changes to:

The radii of two spherical conductors $A$ and $B$ are in the ratio $3:5$ Conductor $\text{'}A\text{'}$ is in air while $B$ is surrounded by a meditm of relative permittivity $6$. The ratio of the capacitances of $A$ and $B$ is

A parallel plate air condenser has a capacity of $20 \mathrm{\mu F}$. What will be the new capacity in $\mathrm{\mu F}$ if a marble slab of dielectric constant $8$ is introduced between the two plates ?

A glass slab is put with in the plates of a charged parallel plates condenser (not connected to battery), which of the following quantities does not change.

If a dielectric substance is introduced between the plates of an isolated charged air-gap capacitor, the energy of capacitor will:-

Two dielectric slab of dielectric constant $K_1$ and $K_2$ and of same thickness is inserted in parallel plates capacitor and $K_1=2K_2$. Potential difference across slabs are $V_1$ and $V_2$ respectively then :-

The potential energy stored in the region between plates of a capacitor is ${\mathrm{U}}_0$. If a dielectric slab of ${\in }_{\mathrm{r}}$ now fills the space between the plates completely then new potential energy will be

The plates of parallel plate capacitor is connected by a battery of emf $\mathrm{ }{\mathrm{V}}_0\mathrm{ }$. The plates are lowered into a vessel containing a dielectric liquid with constant velocity $\mathrm{v}$. Then -