The net Coulomb force on a charge due to multiple charges is the vector sum of forces from each individual charge: F_net = Σ F_i. Holds for both discrete and continuous charge distributions.
-- NCERT Class 12 Physics, Ch. 1, p. 11Superposition Principle
8 MCQs9-step worked example
Source: NCERT ElectrostaticsOfficial key: NTA-verifiedLast updated: 21 Sep 2026
Superposition Principle, explained for NEET
The trap: adding the sizes of the individual forces as plain numbers. Forces from several charges add as vectors. Two 0.90 N forces can give 1.8 N, about 1.6 N, 0.90 N or zero, depending on the angle between them.
The principle. NCERT Class 12 Physics Part I, Chapter 1, Section 1.6 (page 11): experimentally, the force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected by the presence of the other charges. This is the principle of superposition.
For three charges, the total force on q₁ is (NCERT Chapter 1, Eq. 1.4, page 12):
F₁ = F₁₂ + F₁₃
and for n charges it is the vector sum of F₁₂, F₁₃, …, F₁ₙ (Eq. 1.5, page 12). Each term is simply the Coulomb force between that one pair, calculated as if the other charges were absent. The vector sum is taken by the parallelogram law. NCERT adds (page 12) that all of electrostatics is basically a consequence of Coulomb's law and the superposition principle.
Method that avoids the trap.
- Take one source charge at a time; find the size of its force on the chosen charge.
- Fix the direction: like charges push the chosen charge away from the source; unlike charges pull it toward the source.
- Add the vectors: same line → add or subtract with signs; at right angles → √(F₁² + F₂²); two equal forces F at angle θ → 2F cos(θ/2).
- Look for symmetry first. Equal charges placed symmetrically around a point can give a zero total force (NCERT Example 1.5, pages 12–13).
Bridge to NEET. Charges on a line, a right angle, a triangle or a square; find the net force on one. Typical wrong options: magnitudes added as scalars, one force drawn in the wrong direction, or one charge left out.
Watch out: use the angle between the force arrows drawn at the charge, not just an angle of the figure.
Can you answer these Superposition Principle MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
According to NCERT's statement of the principle of superposition, the force on a charge due to several other charges is:
Show answer and why every option is right or wrong
Answer: C. NCERT Class 12 Physics Part I, Chapter 1, Section 1.6 (page 11): the force on any charge due to a number of other charges is the vector sum of the forces due to each, taken one at a time, and the individual forces are unaffected by the other charges.
Why A is wrong: A is wrong because forces have direction; adding magnitudes as plain numbers ignores the angle between them, which is the trap this lesson names.
Why B is wrong: B is wrong because every charge contributes its own Coulomb force; NCERT states the individual forces are unaffected by the presence of other charges, so no charge shields another.
Why D is wrong: D is wrong because NCERT says each individual force is unaffected by the other charges; the pair force is exactly the two-charge Coulomb force, not a reduced one.
MCQ 2Easy RecallPractice
NCERT Chapter 1 states that all of electrostatics is basically a consequence of Coulomb's law and:
Show answer and why every option is right or wrong
Answer: C. NCERT Class 12 Physics Part I, Chapter 1, Section 1.6 (page 12): "All of electrostatics is basically a consequence of Coulomb's law and the superposition principle."
Why A is wrong: A is wrong because conservation of charge is a basic property of charge discussed earlier in the chapter; it is not what NCERT pairs with Coulomb's law in this statement.
Why B is wrong: B is wrong because NCERT uses the consistency of Coulomb's law with Newton's third law to explain why the forces in its three-charge example add to zero; it is not the second half of this statement.
Why D is wrong: D is wrong because quantisation of charge is also a basic property of charge; the statement on page 12 names the superposition principle.
MCQ 3Easy RecallPractice
Three point charges q₁, q₂ and q₃ are held fixed. The force F₁₂ exerted on q₁ by q₂ is:
Show answer and why every option is right or wrong
Answer: B. NCERT Chapter 1, Section 1.6 (page 11): F₁₂ is given by Coulomb's law even though other charges are present; the other charges change only the total force, through the vector sum F₁ = F₁₂ + F₁₃ (Eq. 1.4, page 12).
Why A is wrong: A is wrong because the pair force does not share or split among charges; each pair force is unaffected by the presence of other charges.
Why C is wrong: C is wrong because q₃ adds its own force F₁₃ to the total; it does not change the size of F₁₂.
Why D is wrong: D is wrong because a charge placed between q₁ and q₂ does not block the force of q₂; F₁₂ keeps its Coulomb value and F₁₃ is simply added to it.
MCQ 4Direct ApplicationPractice
A charge of +1.0 μC is at the origin. A charge of +4.0 μC is at x = +0.20 m and a charge of +2.0 μC is at x = −0.10 m. Take 1/(4πε₀) = 9.0 × 10⁹ N m² C⁻². The net force on the +1.0 μC charge is:
Show answer and why every option is right or wrong
Answer: D. Force from +4.0 μC = 9.0 × 10⁹ × (1.0 × 10⁻⁶)(4.0 × 10⁻⁶)/(0.20)² = 0.90 N, repulsive, so along −x. Force from +2.0 μC = 9.0 × 10⁹ × (1.0 × 10⁻⁶)(2.0 × 10⁻⁶)/(0.10)² = 1.8 N, repulsive, so along +x. Vector sum = 1.8 − 0.90 = 0.90 N along +x (superposition, NCERT Chapter 1, Eq. 1.4, page 12).
Why A is wrong: A is wrong because 2.7 N = 1.8 + 0.90 adds the two magnitudes; the forces act in opposite directions along the x-axis, so they must be subtracted.
Why B is wrong: B is wrong because 0.90 N along −x comes from drawing both forces as attractive: the +2.0 μC charge would then pull 1.8 N along −x and the +4.0 μC charge 0.90 N along +x. Like charges repel, so both directions are reversed.
Why C is wrong: C is wrong because 1.8 N is the force of the +2.0 μC charge alone; the 0.90 N force of the +4.0 μC charge has been left out of the sum.
MCQ 5Direct ApplicationPractice
A charge of +1.0 μC is at the origin. A charge of +3.0 μC is at (0.30 m, 0) and a charge of +4.0 μC is at (0, 0.30 m). Take 1/(4πε₀) = 9.0 × 10⁹ N m² C⁻². The magnitude of the net force on the +1.0 μC charge is:
Show answer and why every option is right or wrong
Answer: A. Force from +3.0 μC = 9.0 × 10⁹ × (1.0 × 10⁻⁶)(3.0 × 10⁻⁶)/(0.30)² = 0.30 N along −x; force from +4.0 μC = 9.0 × 10⁹ × (1.0 × 10⁻⁶)(4.0 × 10⁻⁶)/(0.30)² = 0.40 N along −y. The forces are perpendicular, so the resultant is √(0.30² + 0.40²) = 0.50 N (NCERT Chapter 1, Section 1.6, pages 11–12).
Why B is wrong: B is wrong because 0.10 N = 0.40 − 0.30 subtracts the magnitudes as if the forces were opposite; they are perpendicular.
Why C is wrong: C is wrong because 0.25 is 0.30² + 0.40² = 0.09 + 0.16 with the square root not taken; the resultant is √0.25 = 0.50 N.
Why D is wrong: D is wrong because 0.70 N = 0.30 + 0.40 adds the magnitudes as if the forces were along the same line; they are at 90° to each other.
MCQ 6Direct ApplicationPractice
Charge q₁ experiences a force of 3.0 N when only q₂ is present, and a force of 4.0 N in the same direction when only q₃ is present. All three charges are now held in the same positions together. The force on q₁ is:
Show answer and why every option is right or wrong
Answer: D. Each pair force is unaffected by the other charge (NCERT Chapter 1, Section 1.6, page 11), so F₁ = F₁₂ + F₁₃. Both act in the same direction, so the magnitudes add: 3.0 + 4.0 = 7.0 N.
Why A is wrong: A is wrong because 4.0 N keeps only the larger force, as if q₃ blocked q₂; each pair force is unaffected by the presence of the other charge.
Why B is wrong: B is wrong because 5.0 N = √(3.0² + 4.0²) is the resultant for perpendicular forces; these two forces point in the same direction.
Why C is wrong: C is wrong because 1.0 N = 4.0 − 3.0 treats the forces as opposite; the stem says they act in the same direction.
MCQ 7CalculationPractice
Three charges, each +1.0 μC, are at the vertices A, B and C of an equilateral triangle of side 0.10 m in vacuum. Take 1/(4πε₀) = 9.0 × 10⁹ N m² C⁻². The magnitude of the net force on the charge at A is about:
Show answer and why every option is right or wrong
Answer: B. Each pair force = 9.0 × 10⁹ × (1.0 × 10⁻⁶)²/(0.10)² = 0.90 N. At A, the force from B points along BA (away from B) and the force from C along CA (away from C); the angle between them is 60°. Resultant = 2F cos 30° = √3 × 0.90 ≈ 1.6 N, directed away from side BC (NCERT Chapter 1, Section 1.6, parallelogram law, page 12).
Why A is wrong: A is wrong because 1.8 N = 0.90 + 0.90 adds the two magnitudes as if both forces were along the same line; they are 60° apart.
Why C is wrong: C is wrong because 0.90 N is the resultant of two equal 0.90 N forces 120° apart, which is what you get if one force is drawn toward its source charge (attractive) instead of away from it; with two like-charge repulsions the angle is 60°.
Why D is wrong: D is wrong because it copies the zero result of the centroid case; the charge at A is at a vertex, where the two forces are not opposite and do not cancel.
MCQ 8Concept TrapPractice
Four equal positive charges +q sit at the corners A, B, C and D of a square, with A and C diagonally opposite. A positive charge Q is at the centre. Let F be the magnitude of the force on Q due to any one corner charge. The charge at A is now removed. The net force on Q is:
Show answer and why every option is right or wrong
Answer: A. With all four charges present, opposite pairs cancel. After A is removed, the forces from B and D are still equal and opposite and cancel. The force from C is left unbalanced; C repels Q, pushing it away from C, i.e. toward A. Net force = F toward A (superposition, NCERT Chapter 1, Section 1.6, pages 11–12).
Why B is wrong: B is wrong because removing A breaks the symmetry: the force from C no longer has an opposite partner, so the total is not zero.
Why C is wrong: C is wrong because it draws the force of C on Q as attractive (toward C); Q and the charge at C have the same sign, so the force pushes Q away from C, toward A.
Why D is wrong: D is wrong because 3F adds the magnitudes of the three remaining forces; the forces from B and D point in opposite directions and cancel, leaving only F.
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How do you solve a Superposition Principle question? A worked example
- 1
Given
Three charges q₁, q₂, q₃, each equal to q, are at the vertices A, B and C of an equilateral triangle of side l. A charge Q, with the same sign as q, is placed at the centroid O of the triangle.
- 2
Required
The total force on Q.
- 3
Concept
Principle of superposition: the total force on Q is the vector sum of the forces due to the charges at A, B and C, each found from Coulomb's law as if the others were absent. Since Q and q have the same sign, each force pushes Q away from the vertex that produces it.
- 4
Formula
Magnitude of each pair force: F = (1/4πε₀) Qq/r², with r = distance from the vertex to O.
Total force: F_total = F₁ + F₂ + F₃ (vector sum, NCERT Eq. 1.5, page 12). - 5
Substitution
Draw AD perpendicular to BC. AD = AC cos 30° = (√3/2) l.
AO = (2/3) AD = (1/√3) l, and by symmetry AO = BO = CO.
So each force has magnitude (1/4πε₀) Qq/(l/√3)². - 6
Calculation
Each force = (1/4πε₀) × 3Qq/l² = 3Qq/(4πε₀ l²).• F₁ (due to q at A) is along AO.• F₂ (due to q at B) is along BO.• F₃ (due to q at C) is along CO.
F₂ and F₃ are equal and 120° apart. By the parallelogram law their resultant has the same magnitude, 3Qq/(4πε₀ l²), and points along OA.
F₁ has magnitude 3Qq/(4πε₀ l²) along AO, which is exactly opposite to OA.
F_total = [3Qq/(4πε₀ l²)] (r̂ − r̂) = 0, where r̂ is the unit vector along OA.
The factors 3, 2/3, √3, √3/2, cos 30° and the 120° angle are exact geometric numbers; they do not contribute to any significant-figure count. The answer is symbolic, so no rounding is involved. - 7
Final answer
The total force on Q at the centroid is zero.
NCERT also gives a symmetry check: if the resultant were non-zero in some direction, rotating the whole system through 60° about O would give a contradiction (page 13). - 8
Common trap
Adding the three magnitudes and answering 3 × 3Qq/(4πε₀ l²) = 9Qq/(4πε₀ l²). The three forces are 120° apart, so they cancel. A second slip is taking the distance from a vertex to O as l instead of l/√3, which gives each force as Qq/(4πε₀ l²) — the final total is still zero, but any follow-up question that needs a single force will then be wrong by a factor of 3.
- 9
Similar NEET-style question
"In the arrangement of Example 1.5, the charge at vertex A is replaced by −q, while the charges at B and C stay +q and Q stays at the centroid. Find the total force on Q."
Strategy: The charges at B and C still give a resultant of 3Qq/(4πε₀ l²) along OA (toward A). The −q at A now attracts Q, which is also along OA with magnitude 3Qq/(4πε₀ l²). The two add: total = 6Qq/(4πε₀ l²), directed from O toward A. Do not answer zero by copying the original symmetry result.
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What to remember before solving Superposition Principle questions
More in Electrostatics: 3 exam traps and mistakes · 11 formulas · 6 question patterns from its other lessons.
Superposition Principle questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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