Dielectrics play an important role in design of capacitors. The molecules of a dielectric may be polar or non-polar. When a dielectric slab is placed in an external electric field, opposite charges appear on the two surfaces of the slab perpendicular to electric field. Due to this an electric field is established inside the dielectric. The capacitance of a capacitor is determined by the dielectric constant of the material that fills the space between the plates. Consequently, the energy storage capacity of a capacitor is also affected. Like resistors, capacitors can also be arranged in series and/or parallel. (i) Which of the following is a polar molecule ? (ii) Which of the following statements about dielectrics is correct ? (iii) When a dielectric slab is inserted between the plates of an isolated charged capacitor, the energy stored in it : (iv)(a) 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 will become (iv)(b) Two capacitors of capacitances 2\ C_0 and 6\ C_0 are first connected in series and then in parallel across the same battery. The ratio of energies stored in series combination to that in parallel is

PhysicsCBSE Class 12 · Board exam 2024, set 55/5/1 · 4 marks · Case-based

CBSE Class 12 Physics 2024: question 29

Set in the CBSE Class 12 board exam of 2024, paper 55/5/1, section D, question 29, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q29 · 4 marks · Case-basedCompetency-basedElectrostatic Potential and Capacitance

Dielectrics play an important role in design of capacitors. The molecules of a dielectric may be polar or non-polar. When a dielectric slab is placed in an external electric field, opposite charges appear on the two surfaces of the slab perpendicular to electric field. Due to this an electric field is established inside the dielectric.

The capacitance of a capacitor is determined by the dielectric constant of the material that fills the space between the plates. Consequently, the energy storage capacity of a capacitor is also affected. Like resistors, capacitors can also be arranged in series and/or parallel.

(i)
Which of the following is a polar molecule ?
1. (A) $O_{2}$
2. (B) $H_{2}$
3. (C) $N_{2}$
4. (D)HCl
1 mark
(ii)
Which of the following statements about dielectrics is correct ?
1. (A)A polar dielectric has a net dipole moment in absence of an external electric field which gets modified due to the induced dipoles.
2. (B)The net dipole moments of induced dipoles is along the direction of the applied electric field.
3. (C)Dielectrics contain free charges.
4. (D)The electric field produced due to induced surface charges inside a dielectric is along the external electric field.
1 mark
(iii)
When a dielectric slab is inserted between the plates of an isolated charged capacitor, the energy stored in it :
1. (A)increases and the electric field inside it also increases.
2. (B)decreases and the electric field also decreases.
3. (C)decreases and the electric field increases.
4. (D)increases and the electric field decreases.
1 mark
(iv)(a)
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 $(\frac{d}{5})$ is inserted between the plates. The capacitance of the capacitor will become
1. (A) $[\frac{4 K}{5 K + 1}] C_{0}$
2. (B) $[\frac{K + 5}{4}] C_{0}$
3. (C) $[\frac{5 K}{4 K + 1}] C_{0}$
4. (D) $[\frac{K + 4}{5 K}] C_{0}$
1 mark
OR
(iv)(b)
Two capacitors of capacitances $2 C_{0}$ and $6 C_{0}$ are first connected in series and then in parallel across the same battery. The ratio of energies stored in series combination to that in parallel is
1. (A) $\frac{1}{4}$
2. (B) $\frac{1}{6}$
3. (C) $\frac{2}{15}$
4. (D) $\frac{3}{16}$
1 mark

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 55/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

A polar molecule has an unsymmetrical charge distribution.

Hint 2

For an isolated capacitor the charge stays fixed; use U = Q²/2C.

Hint 3

For the slab, use C = ε₀A/(d − t + t/K); for the same battery, energy is proportional to C.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) HCl has an unsymmetrical charge distribution (Cl is more electronegative than H), so it has a permanent dipole moment. O₂, H₂ and N₂ are symmetric.
Option (D).
1 mark
(ii) In an external field the induced dipoles line up along the field, so their net dipole moment is along $E$ .
Option (A) is wrong because without a field the net dipole moment is zero (random orientation). Option (C) is wrong because dielectrics have no free charges. Option (D) is wrong because the induced field opposes the external field.
Option (B).
1 mark
(iii) The capacitor is isolated, so $Q$ stays constant while $C$ becomes $K C$ .
$U = \frac{Q ^{2}}{2 C}$ decreases, and $E = \frac{E _{0}}{K}$ also decreases.
Option (B).
1 mark
(iv)(a) $C = \frac{ε _{0} A}{d - t + t / K}$ with $t = \frac{d}{5}$ :
$C = \frac{ε _{0} A}{d ( \frac{4}{5} + \frac{1}{5 K} )} = [\frac{5 K}{4 K + 1}] C_{0}$
Option (C).
(iv)(b) $C_{s} = \frac{2 C _{0} \times 6 C _{0}}{8 C _{0}} = 1.5 C_{0}$ and $C_{p} = 8 C_{0}$ .
At the same voltage, $\frac{U _{s}}{U _{p}} = \frac{C _{s}}{C _{p}} = \frac{1.5}{8} = \frac{3}{16}$
Option (D).
1 mark

Answer: (i) D; (ii) B; (iii) B; (iv)(a) C; (iv)(b) D

Where the marks are

(i) HCl – D (1 mark)
(ii) B (1 mark)
(iii) B (1 mark)
(iv)(a) 5K C₀/(4K+1) – C, or (iv)(b) 3/16 – D (1 mark)

Revise the chapter

Chapter: Electrostatic Potential and Capacitance (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Electrostatic Potential and Capacitance, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 55/5/1: every question, by section.

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CBSE Class 12 Physics 2024: question 29

Hints

Step-by-step solution

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