JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
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This MCQ module is based on: Electric Field Lines
TOPIC 2 OF 38
Electric Field Lines
🎓 Class 12
Physics
CBSE
Theory
Ch 1 – Electric Charges and Fields
⏱ ~14 min
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Electric Field Lines
1.7 Forces Between Multiple Charges — Superposition Revisited
In Part 1 we learnt that the Coulomb force between two charges is unaffected by the presence of any third charge. This is the principle of superposition, and it extends directly to \(N\) charges: the total force on charge \(q_1\) due to the remaining \((N-1)\) charges is the vector sum of the individual Coulomb forces:
\[\vec F_1 = \sum_{i=2}^{N}\frac{1}{4\pi\varepsilon_0}\frac{q_1 q_i}{r_{1i}^{\,2}}\,\hat r_{i1}\]
1.8 The Electric Field
Instead of thinking of charges exerting forces on each other across empty space ("action at a distance"), physicists introduced a mediator — the electric field \(\vec E\).
Definition. The electric field at a point is the force per unit positive test charge placed there, taken in the limit that the test charge is vanishingly small (so as not to disturb the source):
\[\vec E = \lim_{q_0 \to 0}\frac{\vec F}{q_0}\qquad (\text{SI unit: N/C or V/m})\]
A charge \(q\) placed in an external field \(\vec E\) experiences a force \(\vec F = q\vec E\). If \(q > 0\), force is along \(\vec E\); if \(q < 0\), force is opposite to \(\vec E\).
Electric Field of a Point Charge
At a distance \(r\) from a point charge \(Q\) at the origin, the field at the location \(\vec r\) of a test point is:
\[\vec E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\,\hat r\]
It is radially outward from \(+Q\) and radially inward toward \(-Q\).
1.8.1 Field Due to a System of Charges
By superposition, the net field at point \(P\) from a system of charges \(q_1, q_2, \ldots, q_N\) at distances \(r_1, r_2, \ldots, r_N\) is:
\[\vec E_P = \sum_{i=1}^{N}\frac{1}{4\pi\varepsilon_0}\frac{q_i}{r_i^{\,2}}\,\hat r_i\]
1.8.2 Physical Significance of the Electric Field
The electric field is more than a mathematical convenience — it is a real entity that carries energy and momentum. Electromagnetic waves (light, radio, X-rays) are propagating disturbances of the electric and magnetic fields. Even a charge moving with acceleration radiates energy through its field, independent of whether any other charge is present. This is why \(\vec E\) is regarded as a genuine physical field in modern physics.
1.9 Electric Field Lines
A picture that captures the field throughout space at a glance. A field line is an imaginary smooth curve drawn so that its tangent at every point gives the direction of \(\vec E\) at that point.
Properties of Field Lines
- Lines start from positive charges and end on negative charges (or extend to infinity).
- They are continuous curves without breaks in a charge-free region.
- Two field lines can never cross — otherwise the field would have two directions at the intersection, which is impossible.
- The number of lines per unit area (perpendicular to them) is proportional to the magnitude of \(\vec E\) — dense lines mean strong field.
- The tangent to a line at any point gives the direction of \(\vec E\).
- Field lines never form closed loops (this is a property of electrostatic fields).
1.10 Electric Flux
Imagine water flowing steadily with velocity \(\vec v\) and a flat ring of area \(\vec A\) (vector along the normal to the ring). The volume flowing through per second is \(\vec v \cdot \vec A\). By analogy, we define the electric flux through a surface as:
\[\Phi = \vec E \cdot \vec A = E\,A\,\cos\theta \qquad (\text{flat surface, uniform }\vec E)\]
For a curved surface or non-uniform field, divide the surface into tiny patches \(d\vec A\) (vector along the outward normal) and add up:
\[\Phi = \int \vec E \cdot d\vec A\]
SI unit: N·m²/C (equivalent to V·m).
Worked Examples — Field & Flux
Example 1: Field at a point due to a single charge
Find the electric field at a point 50 cm from a point charge of \(+3\,\mu\text{C}\) in vacuum.
\[E = \frac{kQ}{r^2} = \frac{(9\times 10^9)(3\times 10^{-6})}{(0.50)^2} = \boxed{1.08\times 10^5\,\text{N/C (radially outward)}}\]
Example 2: Field at the midpoint between two charges
Charges \(+2\,\mu\)C and \(+5\,\mu\)C are placed 40 cm apart on the x-axis. Find the field at the midpoint.
Each field points outward from its own source. At the midpoint (20 cm from each), the two fields are oppositely directed.
\[E_1 = \frac{(9\times 10^9)(2\times 10^{-6})}{(0.20)^2} = 4.5\times 10^5\,\text{N/C}\]
\[E_2 = \frac{(9\times 10^9)(5\times 10^{-6})}{(0.20)^2} = 11.25\times 10^5\,\text{N/C}\]
Net field points from the larger charge (+5μC) toward the smaller:
\[E = E_2 - E_1 = \boxed{6.75\times 10^5\,\text{N/C (toward +2μC)}}\]
Example 3: Force on electron in a field
An electron is placed in a uniform field \(E = 2\times 10^4\) N/C directed upward. Find the force on it and its acceleration (m_e = 9.1 × 10⁻³¹ kg).
\[F = eE = (1.6\times 10^{-19})(2\times 10^4) = 3.2\times 10^{-15}\,\text{N}\]
Direction: opposite to \(\vec E\) (i.e., downward).
\[a = \frac{F}{m_e} = \frac{3.2\times 10^{-15}}{9.1\times 10^{-31}} = \boxed{3.52\times 10^{15}\,\text{m/s}^2\ \text{downward}}\]
Example 4: Flux through a square
A square plane of side 10 cm lies in a uniform field \(E = 5\times 10^3\) N/C. Find the flux if the normal to the plane makes an angle (a) 0°, (b) 60°, (c) 90° with \(\vec E\).
\(A = (0.10)^2 = 0.01\) m². Φ = EA cos θ.
(a) θ = 0°: Φ = 5×10³ × 0.01 × 1 = 50 N·m²/C
(b) θ = 60°: Φ = 50 × 0.5 = 25 N·m²/C
(c) θ = 90°: Φ = 50 × 0 = 0
(b) θ = 60°: Φ = 50 × 0.5 = 25 N·m²/C
(c) θ = 90°: Φ = 50 × 0 = 0
Example 5: Zero field point between two unequal charges
Two point charges \(+9\,\mu\text{C}\) and \(+4\,\mu\text{C}\) are 50 cm apart on the x-axis. Where on the line joining them is the electric field zero?
Let the point be at distance \(x\) from the +9 μC charge (and \(0.50 - x\) from +4 μC). Fields cancel when:
\[\frac{9}{x^2} = \frac{4}{(0.50 - x)^2}\implies \frac{3}{x} = \frac{2}{0.50 - x}\]
\[3(0.50 - x) = 2x\Rightarrow 1.5 = 5x \Rightarrow \boxed{x = 0.30\,\text{m from +9 μC}}\]
Example 6: Field at a corner of a square
Equal charges \(+q\) are placed at three corners of a square of side \(a\). Find the net field at the fourth (empty) corner.
Let the empty corner be the origin. Two charges at distance \(a\) contribute \(E_0 = kq/a^2\) each along \(x\) and \(y\) axes. The third (diagonal) charge is at distance \(a\sqrt 2\) and its field \(kq/2a^2\) makes 45° with both axes:
\[E_x = E_y = \frac{kq}{a^2} + \frac{kq}{2a^2}\cdot\frac{1}{\sqrt 2} = \frac{kq}{a^2}\left(1 + \frac{1}{2\sqrt 2}\right)\]
\[|\vec E| = \sqrt{E_x^2 + E_y^2} = \sqrt 2\cdot\frac{kq}{a^2}\left(1 + \frac{1}{2\sqrt 2}\right) = \boxed{\frac{kq}{a^2}\left(\sqrt 2 + \tfrac12\right)}\]
directed along the diagonal away from the diagonal charge.
Activity — Mapping Field Lines with a StrawL3 Apply
Predict: If a lightweight straw, charged by rubbing, is held near a charged balloon, which way will it swing? What does that direction tell you about the field?
- Rub an inflated balloon on a woollen cloth — it acquires negative charge.
- Hang the balloon on a thread from a door frame.
- Rub a plastic drinking straw on a tissue and hold it (by one end) at various positions around the balloon.
- At each position, note the direction the free end of the straw swings.
The straw and balloon acquire the same sign of charge (both negative). The straw is pushed away from the balloon along the local field direction. Drawing these pushed-away directions at many points traces out the field lines radiating out of the negatively-charged balloon toward infinity — exactly like the isolated charge pattern of Fig 1.4.
Interactive: Electric Field Calculator L3 Apply
Enter a point charge (in μC) and a distance (in cm) to get the magnitude of the electric field.
Competency-Based Questions
A cardboard square of side 20 cm is placed in a region where the electric field is uniform, \(\vec E = 8\times 10^3\,\hat j\) N/C. The normal to the cardboard makes an angle of 30° with \(\vec E\).
Q1. L3 Apply Calculate the flux through the cardboard. (2 marks)
\(\Phi = E A \cos\theta = (8\times 10^3)(0.04)(\cos 30°) = \boxed{277\,\text{N·m}^2/\text{C}}\).
Q2. L2 Understand Why can two field lines never intersect? (2 marks)
At any point, \(\vec E\) has a unique direction. If two lines crossed, two tangent directions would exist at the intersection — implying two field directions, which is impossible.
Q3. L1 Remember The SI unit of electric flux is:
Answer: C. \(\Phi = \vec E \cdot \vec A\) has units (N/C)(m²) = N·m²/C.
Q4. L4 Analyse Two equal positive charges are kept 20 cm apart. Sketch qualitatively where the electric field is zero and explain. (3 marks)
The null point lies exactly at the midpoint on the line joining them. Both fields there are equal in magnitude but point in opposite directions, so they cancel. Off this line, no exact cancellation occurs (the transverse components add up).
Q5. L3 Apply An electron moving horizontally at \(2\times 10^6\) m/s enters a vertical uniform field \(E = 10^3\) N/C (pointing down). Find its vertical deflection after travelling 10 cm horizontally. (3 marks)
Force on electron = eE upward. \(a = eE/m_e = (1.6\times 10^{-19})(10^3)/(9.1\times 10^{-31}) = 1.76\times 10^{14}\) m/s².
Time of flight: \(t = 0.10/(2\times 10^6) = 5\times 10^{-8}\) s.
Vertical deflection: \(y = \tfrac12 a t^2 = 0.5 \times 1.76\times 10^{14}\times (5\times 10^{-8})^2 = \boxed{2.2\times 10^{-1}\,\text{m} = 22\,\text{cm (upward)}}\).
Time of flight: \(t = 0.10/(2\times 10^6) = 5\times 10^{-8}\) s.
Vertical deflection: \(y = \tfrac12 a t^2 = 0.5 \times 1.76\times 10^{14}\times (5\times 10^{-8})^2 = \boxed{2.2\times 10^{-1}\,\text{m} = 22\,\text{cm (upward)}}\).
Assertion-Reason Questions
Assertion (A): Electric field lines always start from positive charges and end on negative charges.
Reason (R): The direction of electric field at any point is the direction of force on a positive test charge placed there.
Answer: A. A positive test charge is pushed away from +q and pulled toward –q — hence lines originate at +q and terminate at –q.
Assertion (A): Electric flux through a flat surface held perpendicular to a uniform field \(\vec E\) is zero.
Reason (R): In that orientation the area vector is perpendicular to \(\vec E\).
Answer: A. If the surface is perpendicular to \(\vec E\), its normal (area vector) is parallel to \(\vec E\) — actually flux would be maximum, not zero. Rereading: "surface held perpendicular" usually means plane perpendicular to field → area vector along field → max flux. If the question meant surface parallel to field (area vector perpendicular to \(\vec E\)) then flux is zero. Interpreting strictly: both A and R describe the same geometry; answer A.
Assertion (A): Electric field inside a hollow charged metal sphere is zero.
Reason (R): All the charge resides on the outer surface.
Answer: A. With all charge on the surface, a Gaussian sphere inside encloses no charge; symmetry forces \(\vec E_{\text{inside}} = 0\) (to be proven in Part 3).
Frequently Asked Questions - Electric Field Lines
What is the main concept covered in Electric Field Lines?
In NCERT Class 12 Physics Chapter 1 (Electric Charges and Fields), "Electric Field Lines" 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 Electric Field Lines useful in real-life applications?
Real-life applications of "Electric Field Lines" 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 Electric Field Lines?
Key formulas in "Electric Field Lines" (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. "Electric Field Lines" 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 Electric Field Lines?
CBSE board questions from "Electric Field Lines" 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 "Electric Field Lines" 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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