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?

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?

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

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 ?

What is the effect of dielectric on capacity of a parallel plate capacitor.

Derive an expression for capacity of a parallel plate capacitor filled with dielectric.

On immersing a parallel plate capacitor in oil what will be the effect on its capacitance? The dielectric constant of oil is $2$.

The plate separation of a parallel plate capacitor is $d$. If a metallic plate of thickness $d/2$ is placed in the space between plates not touching any plate, then what will be effect on its capacitance?

On filling the space between the plates of a parallel plate capacitor with some dielectric, its capacitance increases five times. What is the dielectric constant of this material?

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

For capacitor shown in the figure, find capacitance between $A$ and $B$. Area of each plate is $A$ and the plate separation is $d$.

For the capacitor shown in figure, find the capacitance. Area of each plate is $A$, and the plate separation is $d$.

On filling the space between a $2 \mathrm{\mu F}$ air capacitor by mica its capacitance becomes $5 \mathrm{\mu F}$. Calculate the dielectric constant of mica.

A parallel plate capacitor with air as medium is charged to a potential difference $\mathrm{V}$ using a voltage supply. After it is disconnected from the voltage supply, the space between plates is completely filled by a dielectric. Explain with reasons the changes observed in the energy.

A parallel plate capacitor with air as medium is charged to a potential difference $\mathrm{V}$ using a voltage supply. After it is disconnected from the voltage supply the space between plates is completely filled by a dielectric. Explain with reasons the changes observed in the electric field.

A parallel plate capacitor with air as medium is charged to a potential difference $\mathrm{V}$ using a voltage supply. After it is disconnected from the voltage supply the space between plates is completely filled by a dielectric. Explain with reasons the changes observed in the capacitance.