Parallel Plate Capacitor

The two plates of a parallel plate capacitor are $5 \mathrm{mm}$ apart. A slab of a dielectric, of thickness $4 \mathrm{mm}$ is introduced between the plates with its faces parallel to them. The distance between the plates is adjusted so that the capacitance of the capacitor becomes equal to its original value. If the new distance between the plates equal $8 \mathrm{mm}$, what is the dielectric constant of the dielectric used?

On charging a parallel plate capacitor to a potential V, the spacing between the plates is halved, and a dielectric medium of  ${\epsilon }_r=10$  is introduced between the plates, without disconnecting the d.c. source. Explain, using suitable expressions, how the
(i) capacitance,
(ii) potential difference and
​(iii) energy stored in the capacitor change.

If two similar plates, each of area $A$ having surface charge densities  $+\sigma \text{and}-\sigma$ are separated by a distance $d$ in air, write expressions for the potential difference between the plates.

The distance between two plates of a capacitor is $d$ and its capacitance is $C_1$, when air is the medium between the plates. If a metal sheet of thickness $\frac{2d}{3}$ and of the same area as plate is introduced between the plates, the capacitance of the capacitor becomes $C_2$ . The ratio $\frac{C_2}{C_1}$ is

A parallel plate capacitor with plate area $A$ and plate separation $d$ is filled with a dielectric material of dielectric constant $K=4.$ The thickness of the dielectric material is $x,$ where $x<d.$

Let $C_1$ and $C_2$ be the capacitance of the system for $x=\frac{1}{3}d$ and $x=\frac{2d}{3},$ respectively. If $C_1=2 \mathrm{\mu F}$, the value of $C_2$ is _____ $\mathrm{\mu F}.$

As shown in the figure, two parallel plate capacitors having equal plate area of $200 {\mathrm{cm}}^2$ are joined in such a way that $a\ne b$. The equivalent capacitance of the combination is $x{\epsilon }_{0}F$. The value of $x$ is _____.

In a parallel plate capacitor of capacitance $4 \mu \mathrm{F}$, if a dielectric plate ($K=3$) with width equal to half of the width of the capacitor is introduced, the value of capacitance will become :

A metallic slab of thickness $\frac{2d}{3}$ and area of surface same as that of plates of capacitor of capacitance $C_1$ is inserted parallel to plates of capacitor such that its new capacitance becomes equal to $C_2$. If $d$ is the width between the two plates then $\frac{C_2}{C_1}$ is equal to

Area of one side of each plate is $A$ and separation between consecutive plates is$d$. Emf of the battery connected is $V$. The charge on plate $2$ is 

 A slab of material having dielectric constant$K=1.5$ has the same area as the plates of a parallel plate capacitor but has thickness $\frac{3}{4}$ of plate separation introduced between the plates of the capacitor having capacitance $C$. On introducing slab, capacity becomes factor of

A parallel plate capacitor of capacitance $C$ is charged to a potential $V$ by a battery. $Q$ is the charge stored on the capacitor. Without disconnecting the battery, the plates of the capacitor are pulled apart to a larger distance of separation.

What changes will occur in each of the following quantities? Will they increase, decrease or remain the same? Give an explanation in each case.

(a) Capacitance

(b) Charge

(c) Potential difference

(d) Electric field

(e) Energy stored in the capacitor

A capacitor consists of two parallel plates, with an area of cross-section of $0.001 {\mathrm{m}}^2$, separated by a distance of $0.0001 \mathrm{m}$. If the voltage across the plates varies at the rate of ${10}^8 \mathrm{V}/\mathrm{s}$, determine the value of displacement current through the capacitor.

A parallel plate capacitor with plate area $A$ and plate separation$d$, with vacuum between the plates has a capacitance $C_0$.
The space between the plates is filled with two slabs of the same area but thicknesses $\left(\frac{d}{4}\right)$ and $\left(\frac{3d}{4}\right)$ made of material of dielectric constant $2$ and$4$, respectively. The capacitance of the capacitor is now.
 

Four identical metal plates each having area $A,$ are arranged as shown in figure. Find the equivalent capacity of the structure between A and B.
 

Two large parallel conducting plates are given charge $Q_1$ and $Q_2$ respectively as shown in the figure. Find the charge flown through the switch after the switch is closed.

A capacitor of capacitance $C$ consists of two large parallel metal plates. The coefficient of linear expansion of the metal is $\alpha$. What is the change in capacitance if the temperature of the plates rises by $∆T$, while the gap between the plates is kept fixed?

A parallel plate capacitor of capacitance $500\mathrm{pF}$ and plate area $0.05{ \mathrm{m}}^2$ is filled with porcelain $\left(K=6.5\right)$. A potential difference of $200 \mathrm{V}$ is applied. The following observations are recorded. Mark the one which is not correct.

A parallel plate capacitor have capacitance $C$, when area of each plate is halved and distance between plates is doubled, what happened with capacitance when dielectric $2.5$ Introduced between them.

A parallel plate capacitor have capacitance $C$. If area of the plate doubled and distance between plates also doubled, then the capacitance will be

The Plates Of A Parallel Plate Capacitor Are Charged By A Battery And The Battery Is Disconnected After The Charging. Now, The Plates Are Placed As Shown In The Figure. Then (Plates Are Not Parallel To Each Other)