Case Study (Dielectrics): Read the case and answer the sub-questions. (a) Which of the following is a polar mo

Question

Q. Case Study (Dielectrics): Read the case and answer the sub-questions.

(a) Which of the following is a polar molecule?

(i) O

_2

(ii) H

_2

(iii) N

_2

(iv) HCl

(b) Which of the following statements about dielectrics is correct?

(i) A polar dielectric has a net dipole moment in the absence of an external electric field which gets modified due to the induced dipoles.

(ii) The net dipole moment of the induced dipoles is along the direction of the applied electric field.

(iii) Dielectrics contain free charges.

(iv) The electric field produced due to induced surface charges inside a dielectric is along the external electric field.

(c) When a dielectric slab is inserted between the plates of an isolated charged capacitor, the energy stored in it ______.

(i) increases and the electric field inside it also increases

(ii) decreases and the electric field also decreases

(iii) decreases and the electric field increases

(iv) increases and the electric field decreases

(d)

(i) An air-filled capacitor with plate area

A

and plate separation

d

has capacitance

C_0

. A slab of dielectric constant

K

, area

A

and thickness

d/5

is inserted between the plates. The capacitance of the capacitor becomes ______.

(1)

\frac{4K}{5K+1}C_0

(2)

\frac{K+5}{4}C_0

(3)

\frac{5K}{4K+1}C_0

(4)

\frac{K+4}{5K}C_0

(OR)

(d)

(ii) Two capacitors of capacitances

2C_0

and

6C_0

are first connected in series and then in parallel across the same battery. The ratio of energies stored in the series combination to that in the parallel combination is ______.

(1)

\frac{1}{4}

(2)

\frac{1}{6}

(3)

\frac{2}{15}

(4)

\frac{3}{16}

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Part (a)Concept understanding — Dielectrics and Polarisation

A dielectric is an insulating material whose molecules develop a net dipole moment (POLARISATION) when placed in an external field. POLAR molecules (like water) already have a permanent dipole moment due to their asymmetric shape, which an applied field tends to ALIGN; NON-POLAR molecules (like $CO_2$ ) have zero dipole moment normally, but an applied field INDUCES a temporary dipole by pulling their positive and negative charge centres apart. …

Part (b)Concept understanding — Capacitor Energy Storage

Capacitor Energy Storage

The Intuition

Charging a capacitor is like piling sand onto a growing heap. The first grain of charge lands on an empty plate easily. But every later bit of positive charge must be pushed onto a plate that is already positive, and it resists. So more and more work is needed as the plate fills up. All of that work does not disappear — it is stored in the capacitor as electrostatic potential energy, ready to be released later.

Building the Formula

Suppose at some instant during charging the capacitor already holds charge $q$ , so the voltage across it at that moment is $v = q/C$ . Moving one more small charge $dq$ onto the plate costs work:

dW = v\,dq = \frac{q}{C}\,dq

Adding up (integrating) all these small contributions as the charge builds from $0$ to a final value $Q$ gives the total work done:

W = \int_0^{Q} \frac{q}{C}\,dq = \frac{1}{C}\cdot\frac{Q^2}{2} = \frac{Q^2}{2C}

This work is exactly the energy $U$ stored in the charged capacitor.

Three Equivalent Forms

Using $Q = CV$ , the same stored energy can be written three ways — pick whichever matches the quantities you know:

U = \frac{Q^2}{2C} = \frac{1}{2}QV = \frac{1}{2}CV^2

Use $\dfrac{Q^2}{2C}$ when the charge is fixed (capacitor disconnected from the source).
Use $\dfrac{1}{2}CV^2$ when the voltage is fixed (capacitor stays connected to a battery).

Watch out

The factor of $\tfrac{1}{2}$ is essential. A common error is writing $U = QV$ . That would only be true if the full voltage $V$ acted while all the charge moved — but the voltage climbs steadily from $0$ to $V$ as the plates fill, so the effective average voltage is $V/2$ , giving $U = \tfrac{1}{2}QV$ .

Where the Energy Lives — Energy Density

The energy is stored in the electric field occupying the space between the plates, not on the plates themselves. For a parallel-plate capacitor this leads to a general result: energy stored per unit volume of field is

u = \frac{1}{2}\,\varepsilon_0 E^2

where $E$ is the field strength. Wherever an electric field exists, energy is stored there, with density proportional to $E^2$ .

A Quick Example

A $10\ \mu\text{F}$ capacitor is charged to $100\ \text{V}$ . The stored energy is:

U = \frac{1}{2}CV^2 = \frac{1}{2}\times(10\times10^{-6})\times(100)^2 = 0.05\ \text{J}

That 0.05 J can be released almost instantly — which is exactly how a camera flash works: charge slowly, discharge fast. …

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(a) HCl is the polar molecule, the correct dielectric statement is (ii), an isolated capacitor's energy and field both decrease on inserting a slab, and a $d/5$ slab gives $C=\dfrac{5K}{4K+1}C_0$ ; (b) two capacitors $2C_0$ and $6C_0$ store energies in the ratio $\dfrac{U_s}{U_p}=\dfrac{3}{16}$ (series : parallel) across the same battery.

Part (a)

Which molecule is polar? A molecule is polar when the centres of positive and negative charge do not coincide. O $_2$ , H $_2$ and N $_2$ are homonuclear diatomics: the two identical atoms share electrons equally, the charge distribution is symmetric, so there is no permanent dipole. In HCl, chlorine is far more electronegative than hydrogen, so the bonding electrons sit nearer Cl ( $\delta^-$ ) than H ( $\delta^+$ ), giving a permanent dipole moment. Hence the polar molecule is (iv) HCl.
Correct statement about dielectrics. When an external field $\vec E$ is applied, polar molecules rotate to align (and non-polar molecules acquire induced dipoles). In every case the resulting molecular dipole moments point along $\vec E$ , so the material's net dipole moment is along the applied field — statement (ii) is correct. The others are wrong: a polar dielectric has zero net moment in the absence of a field (random orientation); dielectrics contain bound charges, not free charges; and the field of the induced surface charges opposes the external field.
Isolated charged capacitor with a dielectric. "Isolated" means the plate charge $Q$ is fixed (no battery). Inserting a slab of constant $K$ raises the capacitance $C\to KC$ . Then $U=\frac{Q^2}{2C}\;\longrightarrow\;\frac{Q^2}{2KC}=\frac{U}{K},\qquad E=\frac{V}{d}=\frac{Q}{Cd}\;\longrightarrow\;\frac{Q}{KCd}=\frac{E}{K}.$ Both the stored energy and the field decrease (by a factor $K$ ) → (ii). (d)(i) Capacitance with a slab of thickness $d/5$ . The gap is now a dielectric region (thickness $d/5$ ) in series with an air region (thickness $4d/5$ ): $C_1=\frac{K\varepsilon_0A}{d/5}=\frac{5K\varepsilon_0A}{d}=5KC_0,\qquad C_2=\frac{\varepsilon_0A}{4d/5}=\frac{5\varepsilon_0A}{4d}=\frac{5}{4}C_0.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 18 on this concept.

CBSE 2026Set A1 markMCQ
$Q. When a dielectric is placed in an external electric field, the effective electric field inside the dielectric (A) Decreases (B) Increases (C) Becomes zero (D) Remains same$
›Reveal solutionSolution

A dielectric polarises and creates an opposing internal field, so the net field falls to E₀/K.

When a dielectric is placed in an external field $E_0$ , its molecules polarise, developing bound surface charges. These produce an internal field $E_p$ opposite to $E_0$ .

The net (effective) field inside becomes …

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CBSE 2026Set A1 markMCQ
$Q. Dielectric constant of water is (A) 80 (B) 1 (C) 0 (D) \infty$
›Reveal solutionSolution

Water's dielectric constant is about 80.

The dielectric constant (relative permittivity) $\kappa=\varepsilon/\varepsilon_0$ measures how much a medium reduces the electric field between charges compared with vacuum. Water is highly polar, so it is a strong dielectric with $\kappa\approx 80$ . This is why Coulomb forces between ions are …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. In the following dielectric materials are$
(a) Conductor
(b) Super conductor
(c) Semiconductor
(d) Non-conductor
›Reveal solutionSolution

A dielectric is an insulating (non-conducting) material whose molecules get polarised in an external electric field.

Dielectric materials have no free (mobile) charge carriers; under an external electric field their bound charges only shift slightly, creating induced dipoles (polarisation), but no …

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CBSE 2026Set ANNUAL1 mark
$Q. The maximum electric field that a dielectric medium can withstand without break-down is called its ______.$
›Reveal solutionSolution

This maximum field a dielectric can sustain before it breaks down (starts conducting) is called its dielectric strength.

Every insulating (dielectric) material can withstand only up to a certain electric field; beyond that value, the material's bound electrons are ripped free, it starts to conduct, and the insulating property fails (electrical breakdown). This ma …

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CBSE 2026Set ANNUAL1 mark
$Q. What is the dielectric constant of a metal?$
›Reveal solutionSolution

A metal behaves as a conductor with zero internal electric field for any applied field, which is the limiting case K -> infinity in the dielectric-constant formula.

The dielectric constant (relative permittivity) K of a material is defined by K = E0/E, where E0 is the field applied in vacuum and E is the reduced field actually present inside the material once it polarises to oppose the applied field. In a conductor (like a metal), free electrons redistribute instantly so that the NET electric field inside the bulk of the conductor is always exactly zero, however strong the external field E0 is. Putting …

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CBSE 2026Set SEM31 markMCQ
$Q. The capacitances of two conductors X and Y are 3C and C respectively. The conductor X is given a charge Q which the conductor shares with Y. The ratio of total shared energy to its total initial energy is$
(a) 9 : 16
(b) √3 : 2
(c) 3 : 4
(d) 4 : 3
›Reveal solutionSolution

Charge Q on X (capacitance 3C) redistributes over X and Y (total 4C in parallel). Energy = Q²/2C_total, so ratio = (1/4C)/(1/3C) = 3/4 — some energy is lost. Option (c).

Step 1 — initial energy (all charge Q on X, capacitance 3C):

U_i = Q²/(2·3C) = Q²/6C.

Step 2 — after connection, X and Y share charge at a common potential; total capacitance = 3C + C = 4C, and total charge is conserved at Q.

U_f = Q²/(2·4C) = Q²/8C.

…

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CBSE 2025Set A1 markMCQ
$Q. Dielectric constant of air —$
(i) Infinity
(ii) Zero
(iii) One
(iv) Two
›Reveal solutionSolution

Dielectric constant is defined relative to vacuum/air, so air's own dielectric constant is (very nearly) exactly 1.

The dielectric constant (relative permittivity) K of a medium is defined as K = ε/ε₀, the ratio of the permittivity of the medium to the permittivity of free space (vacuum). Since air is very close to vacuum electrically, its permittivity ε_air ≈ ε₀, giving K_air = ε₀/ε₀ = 1 (strictly about 1.0006, taken as 1 for all practical/board purposes). This means air, l …

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CBSE 2024Set A1 markMCQ
$Q. The dielectric constant of a metal is (A) -1 (B) 0 (C) 1 (D) infinity$
›Reveal solutionSolution

A conductor allows charges to move freely, reducing the internal field to zero → dielectric constant → ∞.

The dielectric constant K = E₀/E, where E is the field inside the medium and E₀ the applied field. In a conductor the free charges rearrange so that the internal field E becomes zero.

…

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CBSE 2024Set A1 markMCQ
$Q. The electric dipole moment per unit volume of a substance is called -$
(a) Electric field
(b) Polarisation
(c) Potential
(d) Electric capacitance
›Reveal solutionSolution
Polarisation P is the dipole moment per unit volume induced in a dielectric.

When a dielectric (insulating) material is placed in an external electric field, the field displaces the positive and negative charge centres of each molecule, creating induced dipole moments (or aligning already-present permanent dipoles). The net dipole moment developed per unit volume of the material is called the polarisation, denoted P.

For a linear isotropic dielectric, $P = \chi_e E$ , where $\chi_e$ is the electric susceptibility of the medium and E is the applied field.

Electric field (option a) is force per unit charge, not a dipole-moment density. …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. The value of dielectric strength for air is -$
(a) 3 x 10^6 V/m
(b) 3 x 10^8 V/m
(c) Zero
(d) Infinity
›Reveal solutionSolution

Dielectric strength is the maximum electric field a dielectric can withstand without electrical breakdown; for air this standard value is 3 × 10^6 V/m.

Dielectric strength is defined as the maximum electric field that an insulating (dielectric) medium can tolerate without its molecules ionising and the medium starting to conduct (breaking down). Every dielectric has a characteristic breakdown field. For dry air at atmospheric pressure, this value is a stan …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Dielectric constant for a metal is$
(a) zero
(b) infinite
(c) 1
(d) 10
›Reveal solutionSolution

Dielectric constant K measures how much a material reduces an external field inside it; inside a conductor the net field is always zero, which corresponds to K -> infinity.

For a dielectric placed in an external field $E_0$ , the field inside reduces to $E = E_0/K$ , where K is the dielectric constant. In a conductor (like a metal), free electrons rearrange themselves so as to make the net electric field inside exactly zero in electrostatic equilibrium, i.e. $E = 0$ fo …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The capacity of a condenser is 2\times10⁻⁶ F and its potential is 200 V. The energy released on discharging it fully will be$
(a) 0.02 J
(b) 0.04 J
(c) 0.08 J
(d) 0.16 J.
›Reveal solutionSolution

Energy stored in a charged capacitor is $E = (1/2)CV^2$ ; substituting the given values gives 0.04 J.

When a capacitor of capacitance $C$ is charged to potential difference $V$ , it stores energy

$E = (1/2)CV^2$

This energy is released (as heat/spark) when the capacitor is fully discharged.

Given: $C = 2 \times 10^{-6}$ F, $V = 200$ V.

…

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Q.Case Study (Dielectrics): Read the case and answer the sub-questions.

Part (a)

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