Conductors, Dielectrics, Capacitors and Capacitance – Conductors and insulators: NEET MCQs with Answers

15 free questions on Conductors, Dielectrics, Capacitors and Capacitance – Conductors and insulators for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Conductors and insulators. Dielectrics and electric polarization, capacitors and capacitances, capacitance of a parallel plate capacitor with and without dielectric medium between the plates.

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Question 1

Previous-year question
The capacitance of a parallel plate capacitor with air as the medium is 6 μF. With the introduction of a dielectric medium, the capacitance becomes 30 μF. The permittivity of the medium is (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²) [NEET 2020]
  1. 1.77 × 10⁻¹² C² N⁻¹ m⁻²
  2. 0.44 × 10⁻¹⁰ C² N⁻¹ m⁻²
  3. 5.00 C² N⁻¹ m⁻²
  4. 0.44 × 10⁻¹³ C² N⁻¹ m⁻²
Show answer
Answer: (B) 0.44 × 10⁻¹⁰ C² N⁻¹ m⁻²
Explanation: K = 30/6 = 5, so ε = Kε₀ = 5 × 8.85 × 10⁻¹² = 4.425 × 10⁻¹¹ ≈ 0.44 × 10⁻¹⁰ C² N⁻¹ m⁻².

Question 2

Previous-year question
A parallel plate capacitor of capacitance 20 mF is being charged by a voltage source whose potential is changing at the rate of 3 V s⁻¹. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be respectively [NEET 2019]
  1. zero, 60 mA
  2. 60 mA, 60 mA
  3. 60 mA, zero
  4. zero, zero
Show answer
Answer: (B) 60 mA, 60 mA
Explanation: I = C dV/dt = 20 × 10⁻³ × 3 = 60 mA. Maxwell's displacement current between the plates is always exactly equal to the conduction current in the wires.

Question 3

Previous-year question
The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A is [NEET 2018]
  1. independent of the distance between the plates
  2. linearly proportional to the distance between the plates
  3. inversely proportional to the distance between the plates
  4. proportional to the square root of the distance between the plates
Show answer
Answer: (A) independent of the distance between the plates
Explanation: The force is F = Q²/2Aε₀, which contains no d at all — moving the plates does not change the field one plate creates at the other.

Question 4

A slab of a linear isotropic dielectric of volume 200 cm³ has electric susceptibility χₑ = 3.0. The field inside it is uniform and equal to 2.0 × 10⁵ V/m. What is the net dipole moment of the slab? (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²)
  1. 3.5 × 10⁻¹⁰ C m
  2. 1.1 × 10⁻⁹ C m
  3. 5.3 × 10⁻⁶ C m
  4. 2.7 × 10⁻² C m
Show answer
Answer: (B) 1.1 × 10⁻⁹ C m
Explanation: Polarisation is P = ε₀χₑE = 8.85 × 10⁻¹² × 3.0 × 2.0 × 10⁵ = 5.31 × 10⁻⁶ C m⁻². Since P is dipole moment per unit volume, the net moment is PV. With V = 2.0 × 10⁻⁴ m³, this gives about 1.1 × 10⁻⁹ C m.

Question 5

Consider two statements about a dielectric in an electric field. Statement I: Polarisation is the net dipole moment developed per unit volume of the dielectric. Statement II: Electric susceptibility χₑ, defined for a linear isotropic dielectric by P = ε₀χₑE, has the SI unit C m⁻².
  1. Both statements are true
  2. Statement I is false but Statement II is true
  3. Statement I is true but Statement II is false
  4. Both statements are false
Show answer
Answer: (C) Statement I is true but Statement II is false
Explanation: Dipole moment per unit volume has the unit C m / m³ = C m⁻², and that is what polarisation P means. The product ε₀E also has the unit C m⁻², so χₑ in P = ε₀χₑE must be a pure number. Electric susceptibility is therefore dimensionless and has no SI unit.

Question 6

Two identical air-filled parallel plate capacitors have plate separation d. A copper slab of thickness d/4 is placed between the plates of the first, and a dielectric slab of thickness d/2 and dielectric constant K between the plates of the second, each slab covering the full plate area. If the two capacitances are now equal, K is
  1. 2
  2. 4
  3. 3
  4. 3/2
Show answer
Answer: (A) 2
Explanation: The field inside a conductor is zero, so a copper slab simply reduces the effective air gap to d − d/4 = 3d/4. A dielectric slab of thickness t gives an effective gap of d − t + t/K, which here is d/2 + d/(2K). Equating the gaps gives d/(2K) = d/4, so K = 2.

Question 7

An air-filled parallel plate capacitor with a plate separation of 6 mm is charged to 180 V and disconnected from the battery. A dielectric slab of thickness 2 mm and dielectric constant 3 is then slid in, parallel to the plates. What are the electric field in the remaining air space and the new potential difference between the plates?
  1. 1.0 × 10⁴ V/m and 60 V
  2. 3.0 × 10⁴ V/m and 140 V
  3. 3.0 × 10⁴ V/m and 180 V
  4. 2.0 × 10⁴ V/m and 120 V
Show answer
Answer: (B) 3.0 × 10⁴ V/m and 140 V
Explanation: The charge on the isolated plates does not change, so the field in the air stays at 180 V/6 mm = 3.0 × 10⁴ V/m. Inside the slab the field falls to one third of this, 1.0 × 10⁴ V/m. The new potential difference is (3.0 × 10⁴)(4 × 10⁻³) + (1.0 × 10⁴)(2 × 10⁻³) = 120 V + 20 V = 140 V.
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Question 8

Each molecule of a polar dielectric has a permanent dipole moment. Which statement correctly describes a sample of this dielectric?
  1. It has zero net dipole moment with or without a field, since the dipoles of neighbouring molecules always cancel
  2. It has a large net dipole moment even with no applied field, since every molecule is a dipole
  3. It has zero net dipole moment with no field; a field partly aligns the dipoles along itself, against thermal agitation
  4. It has zero net dipole moment with no field; a field aligns the dipoles opposite to itself, helped by thermal agitation
Show answer
Answer: (C) It has zero net dipole moment with no field; a field partly aligns the dipoles along itself, against thermal agitation
Explanation: With no field, thermal agitation keeps the permanent dipoles randomly oriented, so their total dipole moment is zero. An applied field exerts a torque that turns the dipoles towards its own direction. Thermal motion opposes this alignment, so the result is a net dipole moment along the field that is less than complete alignment would give.

Question 9

An uncharged hollow metal box of irregular shape, with an empty cavity inside, stands near several charged bodies. What is the electric field at points in the cavity once electrostatic conditions are reached?
  1. Zero whatever the shape of the box and whatever the charges outside
  2. Zero only if the box is spherical, since other shapes spread charge unevenly
  3. Non-zero but smaller than outside, since the metal only partly reduces the field
  4. Zero only if the box is connected to the earth, since otherwise charge stays on it
Show answer
Answer: (A) Zero whatever the shape of the box and whatever the charges outside
Explanation: Charges induced on the outer surface of a conductor arrange themselves so that the field inside the metal is zero. With no charge in the cavity, the field in the cavity is also zero, for any shape of conductor and any outside charge arrangement. This is electrostatic shielding, used to protect sensitive instruments from outside electric influence.

Question 10

The rubber tyres of a moving petrol tanker keep its metal body insulated from the road, so a metal chain is often left hanging from the body to touch the ground. What is the main purpose of the chain?
  1. To let charge gathered on the body flow to the earth, so that no spark can form and ignite the fuel
  2. To add a little friction so that the tanker can brake sooner
  3. To keep the centre of gravity of the tanker low on bends
  4. To supply charge from the earth to the body, so that the body stays charged and repels dust
Show answer
Answer: (A) To let charge gathered on the body flow to the earth, so that no spark can form and ignite the fuel
Explanation: Friction during motion can leave charge on the body of the vehicle, and the rubber tyres stop it from leaking away. A conducting path to the ground lets this charge pass into the earth, which is earthing. The body then stays near earth potential, so a spark near the inflammable fuel is avoided.

Question 11

The plates of a parallel plate capacitor carry free charge densities +σ and −σ. The gap is completely filled with a dielectric of dielectric constant 5. What is the magnitude of the bound surface charge density produced on each face of the dielectric?
  1. 0.2σ
  2. 1.25σ
  3. 0.8σ
  4. 5σ
Show answer
Answer: (C) 0.8σ
Explanation: The bound charges on the dielectric faces partly oppose the free charges, so the field inside is E = (σ − σₚ)/ε₀. This field is also 1/K of the value σ/ε₀ without the dielectric. Equating gives σₚ = σ(1 − 1/K) = 0.8σ.

Question 12

A slab of a dielectric made of polar molecules is placed in a uniform external electric field. Which statement best describes what happens in it?
  1. The randomly oriented molecular dipoles partly align with the field, the alignment being opposed by thermal agitation
  2. All the molecular dipoles line up exactly with the field at every temperature
  3. The molecular dipoles turn to point opposite to the field, so the slab gets a net dipole moment against the field
  4. The molecules have no dipole moment until the field is applied, and each then gains one along the field
Show answer
Answer: (A) The randomly oriented molecular dipoles partly align with the field, the alignment being opposed by thermal agitation
Explanation: Polar molecules have permanent dipole moments, but thermal motion keeps them randomly oriented, so the net dipole moment is zero without a field. An external field exerts a torque that tends to align them along the field. Thermal agitation opposes this, so the alignment is only partial, and the slab gains a net dipole moment along the field.

Question 13

In electrostatic equilibrium, why can the electric field at the surface of a charged conductor have no component along the surface?
  1. Because free charges in a conductor can move only at right angles to its surface
  2. Because a component along the surface would make the free charges move until it disappeared
  3. Because the electric field just outside a conductor is always zero
  4. Because the surface charge density must be the same at every point of a conductor
Show answer
Answer: (B) Because a component along the surface would make the free charges move until it disappeared
Explanation: Free charges on a conductor can move along its surface. If the field had a tangential component, it would drive a surface current, and there would be no equilibrium. Charges therefore rearrange until the tangential component vanishes, leaving the field normal to the surface and the surface an equipotential.

Question 14

A parallel plate capacitor filled with a dielectric slab of dielectric constant K stays connected to a battery. The slab is slowly pulled out completely and then slowly put back in its original position. Which statement is correct?
  1. The potential difference across the plates falls to 1/K of its value while the slab is out
  2. The stored energy rises K times while the slab is out and stays at that value after the slab is put back
  3. The charge on the plates stays constant throughout because the battery is never disconnected
  4. The stored energy falls to 1/K of its value while the slab is out and returns to its original value after the slab is put back
Show answer
Answer: (D) The stored energy falls to 1/K of its value while the slab is out and returns to its original value after the slab is put back
Explanation: The battery keeps the potential difference fixed at V. Removing the slab reduces the capacitance to C/K, so some charge flows back to the battery and the energy ½CV² falls to 1/K of its value. Replacing the slab restores C, the charge and the energy, so the net change in stored energy over the whole process is zero.

Question 15

A parallel plate capacitor completely filled with a dielectric slab of dielectric constant 4 has a capacitance of 8 μF. It is charged to 50 V and disconnected from the battery. How much work must an external agent do to pull the slab out slowly and completely?
  1. 30 mJ
  2. 40 mJ
  3. 10 mJ
  4. 20 mJ
Show answer
Answer: (A) 30 mJ
Explanation: The initial energy is ½CV² = ½ × 8 μF × (50 V)² = 10 mJ. The capacitor is isolated, so its charge stays fixed, while its capacitance falls to 2 μF. With U = Q²/2C the energy rises 4 times to 40 mJ. The work done by the external agent equals this rise, 30 mJ.

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