Effect of Dielectric 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?

The two plates of a parallel plate capacitor are 4 mm apart. A slab of dielectric constant 3 and thickness 3 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 2/3rd of its original value. What is the new distance between the plates?

The capacitance of a capacitor is $C_0$. It is connected to a battery of voltage $V$ which charges the capacitor. With the capacitor still connected to the battery, a slab of dielectric material is introduced between the plates of the capacitor. Which of the following explains the effect of the dielectric slab in the above situation?

If a dielectric plate of thickness $\left(t\right)$ is introduced between the plates of a capacitor of plate distance $\left(d\right)$, the capacitor becomes half the original value. Find the dielectric constant of the plate.

In the given figure, a metal slab is inserted in the parallel plates. Which of the option is correct?

$C_P$ (without dielectric) $=2C$

$C_Q$ (without dielectric) $=C$

$k$ and $2k$ are dielectric constant of two dielectrics. Find resultant capacity between $A$ and $B$.

The parallel place capacitor is made of square plates of side length $a$. A parabolic cylindrical slab of dielectric constant $\left(k=2\right)$ is placed between the plates. The equation of parabola is $x^2=ay$ and the vertex of parabola touches the other plate at middle. Then

In diagram shown below, the induced charge on the left face of the dielectric slab of dielectric constant $k$ is (Take area of the left face as $A$)

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

Capacity of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value when a dielectric slab of thickness $t=\frac{d}{3}$ is inserted between the plates of the capacitor of separation$\text{'}d\text{'}.$ Dielectric constant of the slab is $\underline{ }$

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:

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,

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

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 :