JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
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This MCQ module is based on: Dipole Gauss Law
TOPIC 3 OF 38
Dipole Gauss Law
🎓 Class 12
Physics
CBSE
Theory
Ch 1 – Electric Charges and Fields
⏱ ~14 min
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Dipole Gauss Law
1.11 Electric Dipole
An electric dipole is a system of two equal and opposite point charges, \(+q\) and \(-q\), separated by a small distance \(2a\). Many real systems behave as dipoles — polar molecules (H\(_2\)O, HCl), antennas, and the nuclei-electron arrangement in a polarised atom.
Dipole Moment
The dipole moment \(\vec p\) is a vector of magnitude \(p = q(2a)\) directed from \(-q\) to \(+q\).
\[\vec p = q\,(2\vec a),\qquad |\vec p| = 2aq \qquad (\text{unit: C·m})\]
Field on the Axial Line (end-on position)
Consider a point \(P\) on the axis at distance \(r\) from the midpoint \(O\) of the dipole. The distance from +q is \(r-a\) and from –q is \(r+a\). Both fields point along the axis (from +q outward). The net axial field is:
\[E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{(r-a)^2} - \frac{q}{(r+a)^2}\right]
= \frac{1}{4\pi\varepsilon_0}\,\frac{4qar}{(r^2-a^2)^2}\]
For \(r \gg a\) (short dipole limit):
\[\boxed{\ \vec E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\,\frac{2\vec p}{r^3}\ }\]
Direction: along \(\vec p\).
Field on the Equatorial Line (broadside position)
At a point \(Q\) on the perpendicular bisector at distance \(r\) from \(O\), the two individual fields have equal magnitudes \(kq/(r^2+a^2)\). Their components along \(\vec p\) cancel; the components opposite to \(\vec p\) add:
\[E_{\text{eq}} = \frac{1}{4\pi\varepsilon_0}\,\frac{2qa}{(r^2+a^2)^{3/2}}\]
For \(r \gg a\):
\[\boxed{\ \vec E_{\text{eq}} = -\frac{1}{4\pi\varepsilon_0}\,\frac{\vec p}{r^3}\ }\]
Direction: antiparallel to \(\vec p\). Note that the axial field is twice the equatorial field in magnitude and falls off as \(1/r^3\) — faster than a single point charge (\(1/r^2\)) because the two charges nearly cancel at large distances.
1.12 Dipole in a Uniform External Field
Place a dipole in a uniform field \(\vec E\). The two charges experience equal and opposite forces (\(+q\vec E\) on +q; \(-q\vec E\) on –q). The net force is zero, but the forces are not collinear — they form a couple producing a torque.
\[\vec\tau = \vec p \times \vec E,\qquad |\vec\tau| = pE\sin\theta\]
The torque tends to align \(\vec p\) with \(\vec E\):
- \(\theta = 0°\) (p ∥ E): torque = 0, stable equilibrium.
- \(\theta = 180°\) (p antiparallel to E): torque = 0, unstable equilibrium.
- \(\theta = 90°\): torque is maximum = pE.
1.13 Continuous Charge Distribution
When a charge is smeared over a line, surface or volume, we describe it with a charge density.
| Distribution | Density | SI unit | Field element |
|---|---|---|---|
| Line charge | \(\lambda = dq/dL\) | C/m | \(dq = \lambda\,dL\) |
| Surface charge | \(\sigma = dq/dA\) | C/m² | \(dq = \sigma\,dA\) |
| Volume charge | \(\rho = dq/dV\) | C/m³ | \(dq = \rho\,dV\) |
The field of a continuous distribution is obtained by integration:
\[\vec E(\vec r) = \frac{1}{4\pi\varepsilon_0}\int \frac{dq}{r'^{\,2}}\,\hat r'\]
1.14 Gauss's Law
Gauss's Law: The total electric flux through any closed surface is \(1/\varepsilon_0\) times the net charge enclosed by that surface.
\[\oint \vec E \cdot d\vec A = \frac{q_{\text{enc}}}{\varepsilon_0}\]
The closed surface is a mathematical construct called a Gaussian surface. We usually choose it to match the symmetry of the charge distribution so that \(\vec E\) is either perpendicular to \(d\vec A\) (giving zero contribution) or of the same magnitude over the entire surface.
Application 1 — Field of an Infinite Line Charge
Consider an infinite straight wire with uniform linear charge density \(\lambda\). By symmetry, \(\vec E\) is radial (perpendicular to the wire) and has the same magnitude at a given distance. Choose a cylindrical Gaussian surface of radius \(r\) and length \(L\), coaxial with the wire.
- The two flat end caps contribute zero (E is parallel to them).
- The curved side: \(\Phi_{\text{side}} = E \cdot (2\pi r L)\).
- Enclosed charge: \(q_{\text{enc}} = \lambda L\).
Gauss's law gives:
\[E(2\pi r L) = \frac{\lambda L}{\varepsilon_0}\implies \boxed{\ E = \frac{\lambda}{2\pi\varepsilon_0 r}\ }\]
Application 2 — Field of an Infinite Plane Sheet
For a thin sheet with uniform surface charge density \(\sigma\), by symmetry \(\vec E\) is perpendicular to the sheet on both sides. Choose a cylindrical "pillbox" Gaussian surface with flat faces of area \(A\) on either side of the sheet.
- Curved side contributes zero (E parallel to it).
- Two flat faces each contribute \(E A\).
- Enclosed charge: \(\sigma A\).
\[2EA = \frac{\sigma A}{\varepsilon_0}\implies \boxed{\ E = \frac{\sigma}{2\varepsilon_0}\ }\]
Independent of distance! The field near an infinite plane sheet is uniform.
Application 3 — Uniformly Charged Thin Spherical Shell
Let a thin shell of radius \(R\) carry total charge \(Q\) uniformly on its surface. By spherical symmetry, \(\vec E\) is radial and depends only on the distance \(r\) from the centre.
- Outside (r > R): Gaussian sphere of radius \(r\) encloses charge \(Q\). \[E(4\pi r^2) = \frac{Q}{\varepsilon_0}\Rightarrow \boxed{E_{\text{out}} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}}\] The shell acts like a point charge placed at its centre.
- Inside (r < R): Gaussian sphere encloses no charge. \(\boxed{E_{\text{in}} = 0}\)
- At the surface, \(E\) jumps from 0 to \(\sigma/\varepsilon_0\) (where \(\sigma = Q/4\pi R^2\)).
Worked Examples — Dipole & Gauss
Example 1: Dipole moment
Charges +4 nC and –4 nC are separated by 2 mm. Find the dipole moment.
\[p = q(2a) = (4\times 10^{-9})(2\times 10^{-3}) = \boxed{8\times 10^{-12}\,\text{C·m}}\]
directed from –q to +q.
Example 2: Torque on a dipole
A dipole of moment \(p = 5\times 10^{-8}\) C·m is placed in a uniform field \(E = 6\times 10^4\) N/C. The angle between \(\vec p\) and \(\vec E\) is 30°. Find the torque.
\[\tau = pE\sin\theta = (5\times 10^{-8})(6\times 10^4)(\sin 30°) = (5\times 10^{-8})(6\times 10^4)(0.5) = \boxed{1.5\times 10^{-3}\,\text{N·m}}\]
Example 3: Axial field of a short dipole
A short dipole of moment \(p = 2\times 10^{-9}\) C·m is placed along the x-axis. Find the field at \(x = 0.1\) m on the axis.
\[E_{\text{axial}} = \frac{2kp}{r^3} = \frac{2(9\times 10^9)(2\times 10^{-9})}{(0.1)^3} = \boxed{3.6\times 10^4\,\text{N/C}}\]
directed along \(\vec p\).
Example 4: Flux through a sphere
A point charge of \(8\,\mu\)C is placed at the centre of a Gaussian sphere of radius 5 cm. Find the flux through the sphere.
By Gauss's law:
\[\Phi = \frac{q_{\text{enc}}}{\varepsilon_0} = \frac{8\times 10^{-6}}{8.854\times 10^{-12}} = \boxed{9.04\times 10^5\,\text{N·m}^2/\text{C}}\]
The answer is independent of the radius.
Example 5: Field of an infinite line charge
A long straight wire carries a linear charge density \(\lambda = 2\,\mu\)C/m. Find the electric field at a point 20 cm from the wire.
\[E = \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{2\lambda k}{r} = \frac{2(9\times 10^9)(2\times 10^{-6})}{0.20} = \boxed{1.8\times 10^5\,\text{N/C}}\]
directed radially outward.
Example 6: Field due to a charged plane
An infinite non-conducting sheet has surface charge density \(\sigma = 1.7\times 10^{-6}\) C/m². Find the electric field in the region close to the sheet.
\[E = \frac{\sigma}{2\varepsilon_0} = \frac{1.7\times 10^{-6}}{2(8.854\times 10^{-12})} = \boxed{9.6\times 10^4\,\text{N/C}}\]
perpendicular to the sheet on both sides.
Example 7: Charged spherical shell — inside vs outside
A thin spherical shell of radius 10 cm carries a uniform charge of \(+5\,\mu\)C. Find the field at (a) r = 5 cm (inside), (b) r = 15 cm (outside).
(a) Inside a uniformly charged shell, \(\boxed{E = 0}\).
(b) Outside: \[E = \frac{kQ}{r^2} = \frac{(9\times 10^9)(5\times 10^{-6})}{(0.15)^2} = \boxed{2.0\times 10^6\,\text{N/C}}\] radially outward.
(b) Outside: \[E = \frac{kQ}{r^2} = \frac{(9\times 10^9)(5\times 10^{-6})}{(0.15)^2} = \boxed{2.0\times 10^6\,\text{N/C}}\] radially outward.
Activity — Water Stream and a Charged RodL3 Apply
Predict: When a charged plastic rod is brought near a thin stream of water from a tap, what will happen to the stream — and why?
- Open a tap to make a very thin, continuous stream of water.
- Rub a plastic ruler or a balloon with dry wool — it becomes charged.
- Hold the charged object close to (but not touching) the stream.
- Observe which way the water bends.
The water bends toward the charged object — whether the rod is + or –. Water molecules are permanent electric dipoles. The external field of the rod orients each molecule so its opposite charge faces the rod; induced polarisation then attracts the entire stream. This is the molecular origin of why charged objects attract tiny bits of paper, dust, hair — any object containing polar or polarisable material.
Interactive: Gaussian Surface Explorer L3 Apply
Select a geometry, enter the charge/density and distance, and see the electric field formula and value.
Competency-Based Questions
A long straight wire has linear charge density \(\lambda = 4\,\mu\)C/m. A student places a Gaussian cylinder of radius 10 cm and length 50 cm coaxially around it.
Q1. L3 Apply Compute the flux through the cylindrical Gaussian surface. (2 marks)
\(q_{\text{enc}} = \lambda L = (4\times 10^{-6})(0.50) = 2\times 10^{-6}\) C. \(\Phi = q_{\text{enc}}/\varepsilon_0 = 2\times 10^{-6}/8.854\times 10^{-12} = \boxed{2.26\times 10^5\,\text{N·m}^2/\text{C}}\).
Q2. L3 Apply Find the field at 10 cm from the wire. (2 marks)
\(E = \lambda / 2\pi\varepsilon_0 r = 2k\lambda/r = (2\times 9\times 10^9 \times 4\times 10^{-6})/0.10 = \boxed{7.2\times 10^5\,\text{N/C}}\).
Q3. L2 Understand If the wire were replaced by an infinite charged sheet with the same total charge per unit length ≈ the same σ, how would \(E(r)\) change with r? Compare.
Line: \(E \propto 1/r\). Sheet: \(E = \sigma/2\varepsilon_0\), independent of r. The sheet's field extends undiminished, while the line's falls off.
Q4. L1 Remember Which of these is the dipole moment unit?
Answer: C. \(p = q \times (2a)\) has units coulomb times metre = C·m.
Q5. L4 Analyse A dipole of moment 10⁻⁸ C·m is placed in a field 2×10⁵ N/C. Find the work required to rotate it from \(\theta = 0°\) to \(\theta = 90°\). (3 marks)
Potential energy \(U = -pE\cos\theta\).
\(W = U(90°) - U(0°) = -pE(0) - (-pE)(1) = pE = (10^{-8})(2\times 10^5) = \boxed{2\times 10^{-3}\,\text{J}}\).
\(W = U(90°) - U(0°) = -pE(0) - (-pE)(1) = pE = (10^{-8})(2\times 10^5) = \boxed{2\times 10^{-3}\,\text{J}}\).
Assertion-Reason Questions
Assertion (A): Electric field inside a uniformly charged hollow sphere is zero.
Reason (R): A Gaussian sphere drawn inside encloses zero net charge.
Answer: A. By Gauss's law, zero enclosed charge plus spherical symmetry forces \(E = 0\) everywhere inside.
Assertion (A): An electric dipole in a uniform field experiences a net force of zero.
Reason (R): The forces on the two charges are equal in magnitude and opposite in direction.
Answer: A. \(+q\vec E\) and \(-q\vec E\) sum to zero, although they produce a torque.
Assertion (A): The field of an infinite plane sheet is independent of distance from the sheet.
Reason (R): The Gaussian pillbox has flat faces whose area does not change with separation from the sheet.
Answer: A. With the face area A cancelling on both sides of Gauss's law, E comes out as σ/2ε₀ — distance-independent.
Frequently Asked Questions - Dipole Gauss Law
What is the main concept covered in Dipole Gauss Law?
In NCERT Class 12 Physics Chapter 1 (Electric Charges and Fields), "Dipole Gauss Law" 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 Dipole Gauss Law useful in real-life applications?
Real-life applications of "Dipole Gauss Law" from NCERT Class 12 Physics Chapter 1 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 Dipole Gauss Law?
Key formulas in "Dipole Gauss Law" (NCERT Class 12 Physics Chapter 1 Electric Charges and Fields) 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 1?
NCERT Class 12 Physics Chapter 1 (Electric Charges and Fields) is structured so each part builds on the previous one. "Dipole Gauss Law" 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 Dipole Gauss Law?
CBSE board questions from "Dipole Gauss Law" 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 "Dipole Gauss Law" 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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