Dielectrics

A parallel plate air capacitor has capacity 'C' farad, potential 'V' volt and energy 'E' joule. When the gap between the plates is completely filled with dielectric

The function of a dielectric in a capacitor is to

A parallel plate capacitor with air between the plate has a capacitance of $15$ pF. The separation between the plates becomes twice and the space between them is filled with a medium of dielectric constant $3.5$. Then the capacitance becomes $x/4$ pF.

The value of $x$ is

A parallel plate capacitor of plate separation $2$ mm and plate area $40$ cm$2$ is filled with dielectric whose area is same as plates of capacitor. Thickness is $1$ mm and dielectric constant is $5$. New capacitance is ($\epsilon_0$ = permittivity of free space)

A slab of material of dielectric constant $k$ has the same area as the plates of the parallel plate capacitor but has thickness $\frac{3d}{4}$, where $d$ is the separation of the plates. How is the capacitance changed, when the slab is inserted between the plates? ($C$ = original capacitance)

A parallel plate air capacitor consists of two metal plates of area $A$ and separation $d$. A slab of thickness $t$ and dielectric constant $2k$ is inserted between the plates with its faces parallel to the plates. The effective capacitance of the system is ($\epsilon_0 =$ permittivity of free space)

A slab of material of dielectric constant $K$ has the same area as the plates of a parallel plate capacitor but has a thickness $(4/5)d$, where $d$ is the separation of the plates. The capacitance in the presence and absence of dielectric are $C$ and $C_0$ respectively. The ratio $(C/C_0)$ is

A capacitor has capacity C when it's parallel plates are separated by air medium of thickness 'd'. A slab of material of dielectric constant K having area equal to that of plates but thickness $\frac{d}{2}$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be

A parallel plate capacitor has plate area $40$ cm$2$ and plates separation $2$ mm. The space between the plates is filled with a dielectric medium of a thickness $1$ mm and dielectric constant $5$. The capacitance of the system is

A parallel plate capacitor has capacitance $C$, when there is vacuum within the parallel plates. A sheet having thickness $\left(\frac{1}{3}\right)^{\text{rd}}$ of the separation between the plates and relative permittivity $K$ is introduced between the plates. The new capacitance of the system is:

A parallel plate air capacitor has a capacitance $C$. When it is half filled as shown in figure with a dielectric constant $K = 5$, the percentage increase in the capacitance is ________.

Identify the correct statements:

A. Effective capacitance of a series combination of capacitors is always smaller than the smallest capacitance of the capacitor in the combination.

B. When a dielectric medium is placed between the charged plates of a capacitor, displacement of charges cannot occur due to insulation property of dielectric.

C. Increasing of area of capacitor plate or decreasing of thickness of dielectric is an alternate method to increase the capacitance.

D. For a point charge, concentric spherical shells centered at the location of the charge are equipotential surfaces.

Choose the correct answer from the options given below:

A parallel plate capacitor having capacity $12$ pF is charged by a battery of $10$ V between its plates. The battery is now disconnected and a slab of dielectric constant $6$ is kept between the plates. The work done by the capacitor on the slab is

A parallel plate capacitor with plate separation 5 mm is charged by a battery. On introducing a mica sheet of 2 mm and maintaining the connections of the plates with the terminals of the battery, it is found that it draws $25\%$ more charge from the battery. The dielectric constant of mica is ______

In a parallel plate air capacitor of plate separation '$d$', a dielectric slab of thickness '$t$' is introduced between the plates. The capacitance becomes one-third of the original value. The dielectric constant of the slab will be

The capacitance of a capacitor becomes $\frac{7}{6}$ times the original value if dielectric slab of thickness $t = \frac{2d}{3}$ is introduced between the plates, where 'd' is the distance of separation between the plates. The dielectric constant of the slab is

Air capacitor has capacitance $C_1$. The space between two plates of capacitor is filled with two dielectric as shown in figure. The new capacitance of the capacitor is $C_2$. Then $\frac{C_1}{C_2}$ is [$d$ = distance between two plates, $K_1$ and $K_2$ are dielectric constants of first and second dielectrics respectively]

A parallel plate air capacitor has capacity $C$ and plate area $A$. Now the space between its plates is filled by two slabs of thickness $\frac{d}{2}$ each, having dielectric constants $K_1$ and $K_2$. Distance of separation between plates is $d$. The capacity of the capacitor is ($\varepsilon_0$ = permittivity of free space)

Find the capacitance of the Parallel plate capacitor shown in the figure.

(A = area of each plate, $\varepsilon_0$ = permittivity of free space)

Three parallel plate capacitors each with area $A$ and separation $d$ are filled with two dielectrics ( $k_1$ and $k_2$) in the following fashion. Which of the following is true?

$(k_1 > k_2)$

Two parallel plate air capacitors of same capacity '$C$' are connected in series to a battery of emf '$E$'. Then one of the capacitors is completely filled with dielectric material of constant '$K$'. The change in the effective capacity of the series combination is

Two condensers of capacities, '2C' and 'C' are joined in parallel and charged upto potential 'V'. The battery is then disconnected and condenser of capacity 'C' is filled completely with a medium of dielectric constant 'K'. The potential difference across the capacitors in the second case is

Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. 'V'. If one of the capacitor is inserted in liquid of dielectric constant 'K' then potential difference of the other capacitor will become

The space between the plates of a parallel plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants $K_1 = 2, K_2 = 3$ and $K_3 = 5$ (as shown in figure). If new capacitance is $\frac{n}{3}C$ then the value of $n$ is ______.

The function of a dielectric in a capacitor is

A parallel plate air capacitor has capacity ' $C$ ' and distance of separation between plates is ' $d$ '. If a conducting sheet of thickness $\frac{2d}{3}$ is inserted in between the plates, the capacitance becomes $C_1$. The ratio of $\frac{C_1}{C}$ is

A parallel plate capacitor has plate area $50\text{ cm}^2$ and plate separation 3 mm. The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 4. The capacitance becomes ( $\varepsilon_0 =$ permittivity of free space)

The capacity of air filled parallel plate capacitor is $C_0$. One-half of the space between the plates is filled with a dielectric constant ' K ' as shown in figure. The new capacity becomes $C_n$. The ratio $C_n$ to $C_0$ is

A parallel plate air capacitor has capacitance $C_P$. It is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$. Now its capacity becomes $C_K$. The ratio $C_P$ to $C_K$ is

A parallel plate capacitor has charge $5 \times 10^{-6}\text{ C}$. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $4 \times 10^{-6}\text{ C}$ then the dielectric constant of the slab is _______________.

Two parallel plate air capacitors of same capacity ' C ' are connected in parallel to a battery of e.m.f. ' $E$ '. Then one of the capacitors is completely filled with dielectric material of constant ' K '. The change in the effective capacity of the parallel combination is

A parallel-plate capacitor of capacitance $40\mu\text{ F}$ is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant $K = 2$. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are

The plates of a parallel plate capacitor are separated by a distance 'd' with air as the medium between them. A dielectric slab of dielectric constant 3 is introduced between the plates so as to increase the capacity by $50\%$. The thickness of the dielectric slab is

The plates of a parallel plate capacitor are separated by $d$. Two slabs of different dielectric constant $K_1$ and $K_2$ with thickness $\frac{3}{8}d$ and $\frac{d}{2}$, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If $K_1 = 1.25K_2$, the value of $K_1$ is:

A parallel plate capacitor is filled equally(half) with two dielectrics of dielectric constants $\varepsilon_1$ and $\varepsilon_2$, as shown in figures. The distance between the plates is $d$ and area of each plate is $A$. If capacitance in first configuration and second configuration are $C_1$ and $C_2$ respectively, then $\frac{C_1}{C_2}$ is: First Configuration

Second Configuration

Three parallel plate capacitors $C_1, C_2$ and $C_3$ each of capacitance $5\ \mu\text{F}$ are connected as shown in figure. The effective capacitance between points $A$ and $B$, when the space between the parallel plates of $C_1$ capacitor is filled with a dielectric medium having dielectric constant of 4, is:

Space between the plates of a parallel plate capacitor of plate area 4 cm2 and separation of 1.77 mm, is filled with uniform dielectric materials with dielectric constants (3 and 5) as shown in figure. Another capacitor of capacitance 7.5 pF is connected in parallel with it. The effective capacitance of this combination is _ pF.

(Given $\epsilon_0 = 8.85 \times 10^{-12}$ F/m)

The function of dielectric in a capacitor is

A parallel plate capacitor has plate area $40\text{ cm}^2$ and plate separation 2 mm. The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 5. The capacitance of the system is ( $\varepsilon_0 =$ permittivity of vacuum)

A parallel plate capacitor of capacitance $12.5\text{ pF}$ is charged by a battery connected between its plates to potential difference of $12.0\text{ V}$. The battery is now disconnected and a dielectric slab $(\epsilon_r = 6)$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ________ $\times 10^{-12}\text{ J}$.

A parallel plate capacitor with plate separation $5\text{ mm}$ is charged up by a battery. It is found that on introducing a dielectric sheet of thickness $2\text{ mm}$, while keeping the battery connections intact, the capacitor draws $25\%$ more charge from the battery than before. The dielectric constant of the sheet is _________.

Air capacitor has capacitance ' $C_1$ '. The space between two plates of capacitor is filled with two dielectrics as shown in figure. The new capacitance of the capacitor is ' $C_2$ '. The ratio $\frac{C_1}{C_2}$ is $(d =$ distance between two plates of capacitor, $K_1$ and $K_2$ are dielectric constants of two dielectrics respectively)

Air capacitor has capacitance of $1\mu\text{ F}$. Now the space between two plates of capacitor is filled with two dielectrics as shown in figure. The capacitance of the capacitor is [ $d =$ distance between two plates of capacitor, $K_1$ and $K_2$ are dielectric constants of first dielectric and second dielectric respectively]

A parallel plate air capacitor, with plate separation ' d ' has a capacitance of 9 pF. The space between the plates is now filled with two dielectrics, the first having $K_1 = 3$ and thickness $d_1 = d/3$, while the $2$nd has $K_2 = 6$ and thickness $d_2 = 2d/3$. The capacitance of the new capacitor is:

Two identical capacitors A and B are connected in series to a battery of E.M.F., 'E'. Capacitor B contains a slab of dielectric constant $K,\ Q_{A}$ and $Q_{B}$ are the charges stored in A and B. When the dielectric slab is removed, the corresponding charges are $Q'_{A}$ and $Q'_{B}$. Then

Two parallel plate air capacitors of same capacity ' C ' are connected in series to a battery of emf ' $E$ '. Then one of the capacitors is completely filled with dielectric material of constant ' K '. The change in the effective capacity of the series combination is

A capacitor has air as dielectric medium and two conducting plates of area $12\text{ cm}^2$ and they are $0.6\text{ cm}$ apart. When a slab of dielectric having area $12\text{ cm}^2$ and $0.6\text{ cm}$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2\text{ cm}$ to keep the capacitance same as in previous case. The dielectric constant of the slab is: (Given $\epsilon_0 = 8.834 \times 10^{-12}\text{ F/m}$)

A capacitor of $10\ \mu\text{F}$ capacitance whose plates are separated by $10\text{ mm}$ through air and each plate has area $4\text{ cm}^2$ is now filled equally with two dielectric media of $K_1 = 2, K_2 = 3$ respectively as shown in figure. If new force between the plates is $8\text{ N}$. The supply voltage is ________ V.

In a parallel plate capacitor with air between the plates, the distance $d$ between the plates is changed and the space is filled with dielectric constant 8. The capacity of the capacitor is increased 16 times, the distance between the plates is

A parallel plate capacitor with air between the plates has a capacitance of 15pF. The separation between the plates becomes twice and the space between them is filled with a medium of dielectric constant 3.5. Then the capacitance becomes $\frac{x}{4}$ pF. The value of $x$ is ____________.

A parallel plate capacitor is charged by a battery and battery remains connected. The dielectric slab of constant '$K$' is inserted between the plates and then taken out. Then electric field between the plates

The potential energy of charged parallel plate capacitor is $v_0$. If a slab of dielectric constant $K$ is inserted between the plates, then the new potential energy will be

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 has plate area 40 cm$2$ and plates separation 2 mm. The space between the plates is filled with a dielectric medium of a thickness 1 mm and dielectric constant 5. The capacitance of the system is:

A capacitor has capacitance 5$\mu$F when its parallel plates are separated by air medium of thickness d. A slab of material of dielectric constant 1.5 having area equal to that of plates but thickness $\frac{d}{2}$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be __________ $\mu$F.

A parallel plate capacitor with air medium between the plates has a capacitance of $10\mu\text{F}$. The area of capacitor is divided into two equal halves and filled with two media (as shown in figure) having dielectric constant $K_1 = 2$ and $K_2 = 4$. The capacitance of the system will be

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 $C_1 = 2\mu\text{F}$ the value of $C_2$ is __________ $\mu\text{F}$

Two dielectric slabs having dielectric constant '$K_1$ ' and '$K_2$ ' of thickness $\frac{d}{4}$ and $\frac{3d}{4}$ are inserted between the plates as shown in figure. The net capacitance between $A$ and $B$ is $[\varepsilon_0$ is permittivity of free space]

The capacitance of a parallel plate capacitor is $2.5\ \mu\text{F}$. When it is half filled with a dielectric as shown in figure, its capacitance becomes $5\ \mu\text{F}$. The dielectric constant of the dielectric is

Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10\ \mu\text{F}$ are individually charged by a 100 V D.C. source. Capacitor $C_1$ is kept connected to the source and a dielectric slab is inserted between its plates. Capacitor $C_2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor $C_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be ________ V.

(Assuming Dielectric constant $= 10$)

A slab of dielectric constant $K$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4}d$, where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be:

(Given $C_0$ = capacitance of capacitor with air as medium between plates.)

Two metallic plates form a parallel plate capacitor. The distance between the plates is 'd'. A metal sheet of thickness $\frac{d}{2}$ and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?

A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will:

Co is the capacitance of a parallel plate capacitor with air as a medium between the plates (as shown in Fig. 1). If half space between the plates is filled with a dielectric of relative permittivity $\varepsilon$r (as shown in Fig. 2), the new capacitance of the capacitor will be:

A parallel plate capacitor with plate area A and plate separation d = 2 m has a capacitance of 4 $\mu$F. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant K = 3 (as shown in figure) will be:

A composite parallel plate capacitor is made up of two different dielectric materials with different thickness $(t_1$ and $t_2)$ as shown in figure. The two different dielectric materials are separated by a conducting foil $\text{F}$. The voltage of the conducting foil is ____________ V.

A parallel plate capacitor is formed by two plates each of area 30$\pi$ cm2 separated by 1 mm. A material of dielectric strength 3.6 $\times$ 107 Vm${}^{-1}$ is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7 $\times$ 10$-$6C, the value of dielectric constant of the material is:

[Use $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}$ Nm2 C${}^{-2}$]

Two capacitors, each having capacitance $40\ \mu\text{F}$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24\ \mu\text{F}$. The value of $K$ will be:

Two parallel plate capacitors of capacity C and 3C are connected in parallel combination and charged to a potential difference 18 V. The battery is then disconnected and the space between the plates of the capacitor of capacity C is completely filled with a material of dielectric constant 9. The final potential difference across the combination of capacitors will be ___________ V.

A parallel plate capacitor with width $4\text{ cm}$, length $8\text{ cm}$ and separation between the plates of $4\text{ mm}$ is connected to a battery of $20\text{ V}$. A dielectric slab of dielectric constant 5 having length $1\text{ cm}$, width $4\text{ cm}$ and thickness $4\text{ mm}$ is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be ____________ $\epsilon_0$ J. (Where $\epsilon_0$ is the permittivity of free space)

An air filled parallel plate capacitor has a capacity $2\text{ pF}$. The sepeartion of the plates is doubled and the interspace between the plates is filled with dielectric material, then the capacity is increased to $6\text{ pF}$. The dielectric constant of the material is

A parallel plate capacitor filled with oil of a dielectric constant 3 between the plates has capacitance 'C'. If the oil is removed, then the capacitance of the capacitor will be

When the battery across the plates of a charged condenser is disconnected and a dielectric slab is introduced between its plates then the energy stored

A parallel plate capacitor has plate area 100 m2 and plate separation of 10 m. The space between the plates is filled up to a thickness 5 m with a material of dielectric constant of 10. The resultant capacitance of the system is 'x' pF.

The value of $\varepsilon$0 = 8.85 $\times$ 10$-$12 F.m${}^{-1}$.

The value of 'x' to the nearest integer is _____________.

In a parallel plate capacitor set up, the plate area of capacitor is 2 m2 and the plates are separated by 1 m. If the space between the plates are filled with a dielectric material of thickness 0.5 m and area 2 m2 (see fig.) the capacitance of the set-up will be __________$\varepsilon$o. (Dielectric constant of the material = 3.2) (Round off to the Nearest Integer)

If qf is the free charge on the capacitor plates and qb is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge qb an be expressed as:

A parallel plate capacitor of capacitance 200 $\mu$F is connected to a battery of 200 V. A dielectric slab of dielectric constant 2 is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be ____________J.

A parallel plate capacitor whose capacitance C is 14 pF is charged by a battery to a potential difference V = 12 V between its plates. The charging battery is now disconnected and a porcelain plate with k = 7 is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of _____________ pJ. (Assume no friction)

A parallel plate air capacitor is charged upto 100 V. A plate 2 mm thick is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by 1.6 mm. The dielectric constant of the thick plate is

For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4}$ d, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C0) is given by the following relation:

Two parallel plates with dielectric placed between the plates are as shown in figure. The resultant capacity of capacitor will be [A = area of plate. $t_1, t_2$ and $t_3$ are thickness of dielectric slabs, $k_1, k_2$ and $k_3$ are dielectric constants.]

A parallel plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constant K1 and K2 of same area A/2 and thickness d/2 are inserted in the space between the plates. The capacitance of the capacitor will be given by:

In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be: (Given area of plate = A)

Two identical parallel plate air capacitors are connected in series to a battery of emf '$V$'. If one of the capacitor is inserted in liquid of dielectric constant '$K$' then, potential difference of the other capacitor will become

Two capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be: