JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
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This MCQ module is based on: Conductors Dielectrics
TOPIC 6 OF 38
Conductors Dielectrics
🎓 Class 12
Physics
CBSE
Theory
Ch 2 – Electrostatic Potential and Capacitance
⏱ ~14 min
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Conductors Dielectrics
2.8 Electrostatics of Conductors
Metals contain a huge number of free electrons that drift almost instantly when an electric field appears. In electrostatic equilibrium — when everything has settled and no current flows — five important properties follow.
(1) Electric field inside a conductor is zero. If any \(\vec E\) existed inside, free electrons would keep accelerating and we would not be in equilibrium. Charges redistribute on the surface until the interior field becomes \(\vec E=0\).
(2) Net charge lives only on the outer surface. By Gauss's law, any closed surface drawn entirely inside the conductor encloses zero net charge (since \(\vec E=0\) inside). Hence any excess charge must reside on the outer boundary.
(3) Just outside the surface, \(\vec E\) is perpendicular to it. Any tangential component would push surface electrons sideways and drive a current — contradicting equilibrium. Magnitude: \(E = \sigma/\varepsilon_0\), where \(\sigma\) is the local surface charge density.
(4) Potential is the same throughout a conductor. Because \(\vec E=0\) inside and \(\vec E\) is tangential-free on the surface, no work is needed to move a charge from any interior point to any surface point. The entire conductor is an equipotential body.
(5) Electrostatic shielding. The interior of a hollow cavity inside a conductor has \(\vec E=0\), no matter how strong the field outside. This is why sensitive electronics are enclosed in a metal box (Faraday cage) and why cars are generally safe places during a thunderstorm.
2.9 Dielectrics and Polarisation
A dielectric is an insulator that cannot carry steady currents but can respond to an applied electric field at the molecular level. Dielectric molecules fall into two categories.
| Type | Definition | Examples | Behaviour in \(\vec E\) |
|---|---|---|---|
| Non-polar | Centres of positive and negative charge coincide; no permanent dipole moment. | O\(_2\), N\(_2\), CO\(_2\), H\(_2\), CH\(_4\) | Centres shift apart → induced dipole moment parallel to \(\vec E\). |
| Polar | Centres of positive and negative charge are permanently displaced; each molecule is a tiny dipole. | H\(_2\)O, HCl, NH\(_3\), CO | Thermal agitation randomises dipoles; external field partially aligns them along \(\vec E\). |
In both cases the bulk sample develops a polarisation vector \(\vec P\) = dipole moment per unit volume. Inside a polarised slab, bound surface charges appear on its faces; these produce an internal field opposing \(\vec E_{\text{applied}}\), reducing the net field:
\[E_{\text{inside}} = \frac{E_{\text{applied}}}{K}\]
The dimensionless factor \(K\) (also written \(\varepsilon_r\)) is the dielectric constant of the material. Typical values: air \(\approx 1.0006\), paper 3.5, mica 6, glass 5–10, water 80, barium titanate ~1200.
2.10 Capacitors and Capacitance
A capacitor is any pair of conductors separated by an insulator (vacuum, air or a dielectric) and carrying equal but opposite charges \(\pm Q\). Because the conductors are equipotential bodies, they differ by a well-defined potential difference \(V\). Experiment shows that \(V\) is proportional to \(Q\):
Definition. \[C = \frac{Q}{V}\] SI unit: farad (F) = 1 C/V. Practical sub-units: microfarad (1 μF = 10⁻⁶ F), nanofarad (1 nF = 10⁻⁹ F), picofarad (1 pF = 10⁻¹² F).
Crucial property: \(C\) depends only on the geometry (sizes and shapes of the plates, their separation) and the medium between them. Charging the capacitor more does not change \(C\); it only raises \(V\) in direct proportion to \(Q\).
Why capacitance matters. A 1 F capacitor would be the size of a room; real-life devices are μF to pF. Yet these small components store and release energy quickly enough to run every camera flash, smooth every power supply, and tune every radio.
Worked Examples — Conductors, Dielectrics, Capacitance
Example 2.8: Field and charge on an isolated sphere
A copper sphere of radius 10 cm carries a total charge of \(2\,\mu\)C. Find (a) the electric field just outside the surface, (b) the field 5 cm inside the sphere, and (c) the surface charge density.
(a) Just outside: \(E=\dfrac{kQ}{R^2}=\dfrac{(9\times 10^9)(2\times 10^{-6})}{(0.10)^2}=\boxed{1.8\times 10^{6}\,\text{N/C}}\).
(b) Inside a conductor in equilibrium, \(\boxed{E=0}\).
(c) \(\sigma=\dfrac{Q}{4\pi R^2}=\dfrac{2\times 10^{-6}}{4\pi(0.10)^2}=\boxed{1.59\times 10^{-5}\,\text{C/m}^2}\). Check: \(\sigma/\varepsilon_0 = 1.59\times 10^{-5}/8.854\times 10^{-12}=1.8\times 10^{6}\) N/C ✓.
(b) Inside a conductor in equilibrium, \(\boxed{E=0}\).
(c) \(\sigma=\dfrac{Q}{4\pi R^2}=\dfrac{2\times 10^{-6}}{4\pi(0.10)^2}=\boxed{1.59\times 10^{-5}\,\text{C/m}^2}\). Check: \(\sigma/\varepsilon_0 = 1.59\times 10^{-5}/8.854\times 10^{-12}=1.8\times 10^{6}\) N/C ✓.
Example 2.9: Spherical cavity inside a conductor
A hollow metallic sphere of inner radius 5 cm and outer radius 10 cm is given a total charge of \(+6\,\mu\)C. What charge sits on the inner surface, the outer surface, and what is the field inside the cavity?
No charge is placed in the cavity, so Gauss's law on any surface inside the metal gives \(Q_{\text{inner}}=0\). All \(+6\,\mu\)C must sit on the outer surface. Inside the cavity \(\vec E=0\) (electrostatic shielding). \(\boxed{Q_{\text{inner}}=0,\;Q_{\text{outer}}=+6\,\mu\text{C},\;E_{\text{cavity}}=0}\).
Example 2.10: Field reduction by a dielectric
A parallel-plate capacitor is charged to produce a field of \(3\times 10^{5}\) V/m in the gap. A slab of K = 5 is slipped into the gap while the charge on the plates is held fixed. What is the new field inside the dielectric?
\[E_{\text{new}}=\frac{E_0}{K}=\frac{3\times 10^{5}}{5}=\boxed{6\times 10^{4}\,\text{V/m}}\]
The dielectric partially cancels the original field through induced bound charges.
Example 2.11: Capacitance from charge and voltage
When a capacitor receives 12 nC of charge, its plates develop a potential difference of 6 V. Find its capacitance.
\[C=\frac{Q}{V}=\frac{12\times 10^{-9}}{6}=\boxed{2\,\text{nF}}\]
Example 2.12: Polar vs non-polar
Why does water (K ≈ 80) have a much larger dielectric constant than hydrogen gas (K ≈ 1.00026) at room temperature?
Water molecules are polar — each carries a permanent dipole moment ~6.2 × 10⁻³⁰ C·m, which a weak external field can partially align. The resulting bulk polarisation is very large. Hydrogen is non-polar; only tiny induced dipoles (~10⁻³⁵ C·m) form, giving a K barely above 1.
Example 2.13: Charge shared between conductors
Two isolated conductors of capacitances \(C_1=3\,\mu\)F and \(C_2=6\,\mu\)F carry charges \(Q_1=30\,\mu\)C and \(Q_2=0\) respectively. They are connected by a wire. Find the final common potential.
After connecting, both become a single equipotential conductor with total charge \(30\,\mu\)C on total capacitance \((C_1+C_2)=9\,\mu\)F.
\[V_{\text{common}}=\frac{Q_{\text{tot}}}{C_1+C_2}=\frac{30\times 10^{-6}}{9\times 10^{-6}}=\boxed{3.33\,\text{V}}\]
Activity — Home-made Faraday CageL3 Apply
Predict: Wrap a mobile phone completely in aluminium foil, then call it from another phone. Will the call go through? Why?
- Take a fully charged mobile phone; ask a friend to call it — confirm the call rings (baseline).
- Switch off the phone, wrap it snugly in 2–3 layers of aluminium foil, leaving no gap.
- Switch it on (through a tiny flap), wait 30 s and call again from another phone.
- Observe the result and the signal-bar reading.
Observation: The foil-wrapped phone receives no signal; the call goes straight to voicemail.
Explanation: The aluminium foil is a conductor; its free electrons rearrange so quickly that electromagnetic waves trying to enter set up surface currents instead of reaching the interior. The cavity is shielded — a Faraday cage in action. The same principle shields MRI rooms, aircraft cockpits and sensitive electronics.
Explanation: The aluminium foil is a conductor; its free electrons rearrange so quickly that electromagnetic waves trying to enter set up surface currents instead of reaching the interior. The cavity is shielded — a Faraday cage in action. The same principle shields MRI rooms, aircraft cockpits and sensitive electronics.
Interactive: Capacitance Explorer L3 Apply
Vary plate area, separation and dielectric constant to see how the capacitance of a parallel-plate capacitor \(C = K\varepsilon_0 A/d\) responds.
Competency-Based Questions
A laboratory has a solid copper sphere of radius 8 cm mounted on an insulating stand. A second concentric hollow copper shell (inner radius 12 cm, outer 14 cm) surrounds it and is earthed. A charge of \(+3\,\mu\)C is given to the inner sphere.
Q1. L1 Remember What is the electric field at a point 4 cm from the centre (inside the inner sphere)?
Answer: C. Inside any conductor in equilibrium \(E=0\).
Q2. L3 Apply Find the field at a point 10 cm from the centre (between the two conductors). (2 marks)
\(E = \dfrac{kQ}{r^2}=\dfrac{9\times 10^9\times 3\times 10^{-6}}{(0.10)^2}=\boxed{2.7\times 10^{6}\,\text{N/C}}\), directed radially outward.
Q3. L2 Understand Fill in the blank: Because the outer shell is earthed, an induced charge of ____ appears on its inner surface.
\(\boxed{-3\,\mu\text{C}}\) — equal and opposite to the inner sphere's charge. The matching \(+3\,\mu\text{C}\) that would normally sit on the outer surface flows to earth.
Q4. L3 Apply A glass slab with K = 6 is placed in a 200 V/m field. Find the field inside the slab. (2 marks)
\(E_{\text{inside}} = 200/6 = \boxed{33.3\,\text{V/m}}\).
Q5. L4 Analyse True/False: When a capacitor is connected across a fixed battery and a dielectric is inserted, the charge on the plates stays the same.
False. The voltage is held fixed by the battery; with \(C\) rising by factor K, \(Q=CV\) also rises by factor K — extra charge flows from the battery onto the plates.
Assertion-Reason Questions
Assertion (A): The electric field just outside a charged conductor is always normal to the surface.
Reason (R): A tangential component would drive a surface current and destroy electrostatic equilibrium.
Answer: A. The requirement of equilibrium rules out tangential fields.
Assertion (A): Water has a higher dielectric constant than most solids at room temperature.
Reason (R): Water molecules possess a permanent dipole moment which aligns with the applied field.
Answer: A. Orientational polarisation of polar molecules is far stronger than induced polarisation alone.
Assertion (A): Capacitance of a capacitor depends only on its geometry and the medium.
Reason (R): The ratio Q/V is fixed once the shape, size and dielectric are chosen.
Answer: A. Double the charge and the voltage doubles — the ratio is unchanged.
Frequently Asked Questions - Conductors Dielectrics
What is the main concept covered in Conductors Dielectrics?
In NCERT Class 12 Physics Chapter 2 (Electrostatic Potential and Capacitance), "Conductors Dielectrics" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Conductors Dielectrics useful in real-life applications?
Real-life applications of "Conductors Dielectrics" from NCERT Class 12 Physics Chapter 2 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Conductors Dielectrics?
Key formulas in "Conductors Dielectrics" (NCERT Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 2?
NCERT Class 12 Physics Chapter 2 (Electrostatic Potential and Capacitance) is structured so each part builds on the previous one. "Conductors Dielectrics" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Conductors Dielectrics?
CBSE board questions from "Conductors Dielectrics" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Conductors Dielectrics" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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