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?

A parallel plate condenser is charged and isolated. When a sheet of glass is interposed between the plates

A dielectric material gets _____ when it placed in an electric field.

 When a dielectric is placed in an external electric field a dielectric gets polarised.This statement is ture or false.

When dielectric is placed in an electric field the field within the dielectric is?

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?

A dielectric slab fills the lower half region of parallel plate capacitor as shown in figure. Take plate area as $A$ then incorrect statement 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,

The charges on two parallel copper plates separated by a small distance are $+Q$ and $-Q$. A test charge q experiences a force $F$ when placed between these plates. Now if one of the plates is removed to infinity, then the force on the test charge will become-

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

One plate of a capacitor is connected to a spring as shown in figure. Area of both the plates is $A$. In steady state separation between the plates is $0.8d$ (spring was unstretched and the distance between the plates was $d$ when the capacitor was uncharged). The force constant of the spring is approximately:

Dielectric strength of air is $3\times {10}^6 \mathrm{V}/\mathrm{m}.$ What maximum potential upto which a capacitor of capacity $2 \mathrm{\mu F}$ and plate separation is $0.02 \mathrm{mm}$ is charged?

The plates of a parallel plate capacitor are charged up to $100$ volt. A $2 \mathrm{mm}$ thick plate is inserted between the plates, then to maintain
the same potential difference, the distance between the capacitor plates is increased by $1.6 \mathrm{mm}$. The dielectric constant of the plate is :

The force between the plates of a parallel plate capacitor of capacitance $C$ and distance of separation of the plates $d$ with a potential difference $V$ between the plates, is-