Q.Consider three charges q1, q2, q3 each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle?
Concept understanding — Coulomb Force Superposition
Coulomb Force Superposition – From Intuition to Precision
Imagine you're in a room with three friends. Each friend can push or pull you. If two friends push you from the same side, you feel a stronger push — the combined effect. If one pushes from the left and another from the right, you feel the net effect, which might be smaller or even zero if they push equally hard.
This is exactly how electric forces work. When multiple charged particles are present, each one exerts its own force on a given charge. The total force that charge feels is simply the vector sum of all the individual forces — as if each other charge were acting alone, completely ignoring the presence of the rest.
That's the core idea: forces add like arrows, not like numbers.
The Precise Statement
Fnet on q0=∑i=1nFi→0=4πε01∑i=1nri02q0qir^i0
Where:
- q0 is the charge you're calculating the force on
- qi are all other charges (excluding q0 itself)
- ri0 is the distance between qi and q0
- r^i0 is a unit vector pointing from qi to q0 (or away, depending on sign convention — be consistent)
The key point: Each pair of charges interacts independently. The presence of a third charge does not alter the force between the first two. This is what "superposition" means — the forces simply layer on top of each other.
Why This Matters (and a Common Trap)
Never add the magnitudes of forces directly unless all forces are along the same line and in the same direction. Force is a vector — direction matters.
If two forces point in opposite directions, they partially cancel. If they're at right angles, the net force is found using the Pythagorean theorem, not simple addition.
Example: Three charges on a line:
- q1=+2μC at x=0
- q2=−1μC at x=3cm
- q0=+1μC at x=1cm
Step 1: Force from q1 on q0 — both positive, so repulsive. q0 is pushed to the right.
Step 2: Force from q2 on q0 — opposite signs, so attractive. q0 is pulled to the right (toward q2).
Step 3: Both forces point right. Now you add magnitudes: Fnet=F1→0+F2→0.
If q2 were also positive, the force from q2 would push q0 left, and you'd subtract.
The Deeper Reason
Coulomb's law is a linear law — the force is proportional to each charge individually. If you double q1, the force from q1 doubles, but the force from q2 stays the same. This linearity is what makes superposition possible. It's not a coincidence — it's a fundamental property of electromagnetic interactions at the classical level.
Superposition works because electric forces obey a linear inverse-square law. If the force depended on products of three charges (like q0q1q2), superposition would fail. It doesn't — and that's why we can break down any multi-charge problem into a series of two-charge calculations.
How to Use It in Exams
- Draw all charges and label distances.
- For each other charge, sketch the direction of the force on your target charge (like charges repel, opposites attract).
- Write the magnitude of each force using Coulomb's law.
- Resolve into components if forces aren't along the same line.
- Add components separately: Fnet,x=∑Fi,x, same for y, z.
- Combine components to get the net force vector.
In symmetric arrangements (e.g., an equilateral triangle with equal charges), many components cancel. Always check for symmetry before diving into heavy algebra — it can save you minutes.
One Last Check
If you place a test charge q0 at a point and there are 10 other charges around it, you calculate 10 separate Coulomb forces and add them as vectors. That's it. No extra physics, no hidden interactions. The universe, at this level, is beautifully simple: each pair talks only to each other, and you just listen to all the conversations at once.
"Coulomb's law superposition principle examples" and "electrostatics class 12 physics important questions" are frequently searched, both grounded in the Electrostatics chapter of the NCERT/CBSE Class 12 Physics curriculum. Multi-charge force problems using superposition are a near-guaranteed topic in JEE Main and NEET.
Why this formula?
Coulomb Force Superposition — Why the Formula Holds
The principle of superposition for Coulomb forces states that the net electrostatic force on a given charge due to a collection of other charges is the vector sum of the individual forces from each charge, as if the others were absent.
The Key Formula
If we have a charge q0 at position r0, and N other point charges q1,q2,…,qN at positions r1,r2,…,rN, the net force on q0 is:
Fnet=4πε01∑i=1N∣r0−ri∣2q0qir^0i
where r^0i is the unit vector pointing from qi to q0.
Why This Works — The Physical Reasoning
1. Coulomb's Law is a Two-Body Interaction
Coulomb's law describes the force between exactly two point charges. It depends only on:
- The product of their charges (q0qi)
- The inverse square of the distance between them
- The direction along the line joining them
Crucially, the force between q0 and qi does not depend on the presence of any other charges qj.
2. Forces Add as Vectors (Newton's Third Law + Linearity)
Electrostatic forces are real physical forces — they obey Newton's laws. If multiple forces act on the same charge, the net effect is the vector sum of each individual force. This is a fundamental property of forces in classical mechanics.
3. The Electric Field is Linear
A deeper reason: the electric field E obeys superposition. Since F=q0E, and E from multiple sources adds linearly, the force automatically adds linearly.
The electric field at r0 due to qi is:
Ei(r0)=4πε01∣r0−ri∣2qir^0i
Then:
Fnet=q0∑iEi=∑iFi
The Crucial Assumption (Why It's Not Trivial)
Superposition holds because Maxwell's equations are linear in the electric field. If the equations were nonlinear (e.g., if the field depended on E2), then the force from two charges together would not be the sum of the individual forces.
In electrostatics, the electric field satisfies:
∇⋅E=ε0ρ,∇×E=0
Both equations are linear — if E1 and E2 are solutions, then E1+E2 is also a solution. This linearity is the mathematical reason superposition works.
Exam-Relevant Takeaway
| Concept | Why It Holds |
|---|---|
| Superposition of forces | Coulomb force is a two-body interaction; forces add as vectors |
| Superposition of fields | Maxwell's equations are linear in E |
| Net force formula | Fnet=∑Fi — vector sum of individual Coulomb forces |
Never forget: The unit vector r^0i points from the source charge to the test charge — this determines the correct direction of each term.
Quick Example (To Cement the "Why")
Suppose q0=+1μC at the origin, q1=+2μC at (1,0), q2=−2μC at (0,1).
- Force from q1: repulsive, along +x direction
- Force from q2: attractive, along +y direction
The net force is not just the sum of magnitudes — it's the vector sum:
Fnet=F1x^+F2y^
This works because the two forces are independent — q1 doesn't "know" about q2, and vice versa. The superposition principle is simply the statement that this independence holds.
Concept: Coulomb Force Superposition — the net force on Q is the vector sum of three individual repulsive forces from q1,q2,q3.
- Each side of the equilateral triangle is l. The distance from a vertex to the centroid is 3l. So each repulsive force has magnitude:
F=4πε01(l/3)2qQ=4πε03l2qQ
-
The three forces lie along the medians, pointing away from the vertices. At the centroid, the medians are separated by 120∘.
-
Three vectors of equal magnitude, spaced 120∘ apart, sum to zero. This is true regardless of the sign of Q and q (as long as they are the same sign, all forces are either all repulsive or all attractive).
The net force on Q is 0.
The three equal repulsive forces on Q from the three vertices are equal in magnitude and spaced 120∘ apart, so their vector sum is zero. The net force on Q is 0.
The key idea is Coulomb’s law with superposition. Each vertex charge q exerts a repulsive force on Q (since both have the same sign). Because the triangle is equilateral, the centroid is equidistant from all three vertices, so each force has the same magnitude. And because the three vertices are symmetrically placed around the centroid, the three force vectors point along the medians, 120∘ apart. When three equal vectors are arranged at 120∘ intervals, they cancel exactly.
Let’s work through it step by step.
- Distance from centroid to each vertex. In an equilateral triangle of side l, the centroid is also the circumcenter. The distance from the centroid to any vertex is the circumradius:
R=3l.
(Derivation: the altitude is 23l, and the centroid divides each median in the ratio 2:1, so the distance from centroid to vertex is 32 of the altitude: 32⋅23l=3l.)
- Magnitude of each force. By Coulomb’s law, the force on Q due to a single vertex charge q is
F=4πε01R2∣qQ∣=4πε01(l/3)2qQ=4πε01l23qQ.
Since q and Q have the same sign, the force is repulsive — it points directly away from that vertex.
-
Direction of each force.
The centroid lies at the intersection of the medians. The line from a vertex to the centroid is exactly along the median. So the force from vertex A points from O away from A (straight down in the textbook figure), from B away from B (up-right), and from C away from C (up-left). These three directions are separated by 120∘.
-
Vector addition.
Place the three force vectors tail-to-tail at O. They have equal magnitude F and are spaced 120∘ apart. Their resultant is zero.
TipA quick way to see this: the sum of three equal vectors at 120∘ is zero because they form the sides of an equilateral triangle when placed head-to-tail. Alternatively, resolve each into components: the horizontal components cancel pairwise, and the vertical components also sum to zero.
Explicitly, take the direction from O toward A as the negative y-axis. Then:
- FA=−Fj^
- FB=Fsin60∘i^+Fcos60∘j^=23Fi^+21Fj^
- FC=−Fsin60∘i^+Fcos60∘j^=−23Fi^+21Fj^
Adding:
Fnet=(23F−23F)i^+(−F+21F+21F)j^=0i^+0j^=0.
A common mistake is to think the forces cancel only if Q is at the center of the triangle — but that’s exactly the centroid. Another pitfall: forgetting that the forces are repulsive and pointing away from the vertices, not toward them. If you mistakenly draw them pointing inward, they’d add to a nonzero resultant.
The net force on Q is zero: 0.
Instead of resolving each force into components, use a pure symmetry argument: the charge configuration is unchanged by a 120∘ rotation about the centroid, so the net force there must be too — and the only vector unchanged by a 120∘ rotation is the zero vector. Net force =0.
Method: Rotational-Symmetry Argument
This problem can be solved without computing a single force magnitude, just by reasoning about symmetry — often faster and less error-prone than vector addition.
-
Set up the symmetry.
The three charges q1=q2=q3=q sit at the vertices of an equilateral triangle, with Q at the centroid. Rotate the entire triangle by 120∘ about the centroid: vertex 1 moves to where vertex 2 was, vertex 2 to where vertex 3 was, and vertex 3 to where vertex 1 was.
-
Observe that the configuration looks identical after rotation.
Because all three vertex charges are equal (q1=q2=q3=q), swapping their positions this way leaves the physical charge distribution completely unchanged. An observer at the centroid cannot tell the triangle was rotated.
-
The force on Q must obey the same symmetry.
Since the source charges look identical before and after the rotation, the electric force they produce on Q (sitting exactly at the centroid, the rotation axis) must also look identical before and after — i.e., the net force vector F must map onto itself when rotated by 120∘.
-
Ask what vectors are invariant under a 120∘ rotation.
Rotating any nonzero vector by 120∘ always produces a different vector (pointing in a different direction) — 120∘ is neither 0∘ nor a multiple of 360∘. The only vector that is unchanged by such a rotation is the zero vector.
-
Conclude.
Therefore F must equal the zero vector:
Fnet on Q=0
This symmetry method generalizes well: for any n equal charges arranged symmetrically (n≥3) around a central point, the net force or field at the center is zero by the same rotational argument — no need to redo the component algebra for a square, pentagon, or hexagon of equal charges.
The net force on Q is 0.
Here are the most common mistakes students make when solving this classic Coulomb force superposition problem, along with how to avoid each.
1. Forgetting the Vector Nature of Force
The Mistake:
Students often compute the magnitude of the force from each q on Q correctly, but then simply add them as scalars (e.g., Fnet=F1+F2+F3).
Why it’s wrong:
Coulomb force is a vector. Forces from different charges point in different directions. Adding magnitudes directly ignores direction and gives an incorrect (usually larger) result.
How to Avoid:
Always draw a clear diagram showing the direction of each force vector. Use vector addition (component method or symmetry) — never scalar addition.
2. Not Using Symmetry to Simplify
The Mistake:
Students calculate all three force vectors explicitly, resolve into components, and sum — a long, error-prone process.
Why it’s wrong:
It wastes time and increases the chance of algebraic mistakes. The problem has perfect symmetry.
How to Avoid:
Recognize that the three charges are identical and placed at vertices of an equilateral triangle. The centroid is equidistant from all vertices. By symmetry, the three force vectors are equal in magnitude and spaced 120∘ apart. Their vector sum is zero.
Key result: The net force on Q at the centroid is Fnet=0.
3. Incorrect Distance Calculation
The Mistake:
Using l (side length) as the distance between a vertex charge and the centroid.
Why it’s wrong:
The distance from a vertex to the centroid of an equilateral triangle is not l. It is 3l.
How to Avoid:
Memorize or derive:
- Centroid divides the median in ratio 2:1.
- Median length =23l.
- Distance from vertex to centroid =32×median=32⋅23l=3l.
Use r=3l in Coulomb’s law.
4. Sign Confusion in Force Direction
The Mistake:
If Q and q have the same sign, students sometimes draw forces as attractive.
Why it’s wrong:
Like charges repel. All three forces on Q are repulsive and point radially outward from each vertex.
How to Avoid:
Always check: same sign → repulsion (force away from the other charge). Opposite sign → attraction (force toward the other charge). Draw arrows accordingly.
5. Assuming the Net Force is Non-Zero Without Checking
The Mistake:
After computing magnitudes, students assume the forces don’t cancel and proceed to find a non-zero resultant.
Why it’s wrong:
Symmetry guarantees cancellation. The three equal-magnitude vectors at 120∘ to each other always sum to zero.
How to Avoid:
Before doing heavy algebra, pause and check for symmetry. If the configuration is symmetric and all charges are identical, the net force at the center is zero.
Quick Summary Checklist
| Mistake | How to Avoid |
|---|---|
| Scalar addition of forces | Always use vector addition |
| Ignoring symmetry | Use symmetry to simplify first |
| Wrong distance (l instead of l/3) | Derive or memorize centroid distance |
| Wrong force direction (attraction instead of repulsion) | Same sign → repulsion |
| Assuming net force is non-zero | Check symmetry — here it’s zero |
Final takeaway: For this exact problem, the answer is zero — but only if you handle vectors, distances, and directions correctly.
Showing the 12 most recent of 40 on this concept.
- CBSE 2026Set A1 markMCQQ.Coulomb's law is valid for (A) Point charges only (B) Dispersed charges only (C) Both point charges and dispersed charges (D) Neutral particles
›Reveal solutionSolution
Coulomb's law is defined for point charges; extended bodies need integration.
Coulomb's law states F=4πε01r2q1q2, where r is the distance between the charges.
This form requires the charges to be point charges so that a single unambiguous separation r exists. For a continuous/dispersed charge distribution the distance from each element differs, so one must integrate Coulomb's law over the distribution rather than apply it directly.
✓Final answer(A) Point charges only.
- CBSE 2026Set ANNUAL1 markMCQQ.Two spheres carrying charges +6 μC and +9 μC, separated by a distance d, experience a force of repulsion F. When a charge of −3 μC is added to each sphere and distance d is kept the same, the new force of repulsion will be(a) 3F(b) F/9(c) F(d) F/3
›Reveal solutionSolution
New force = F/3, because force is proportional to the product of the charges and only the charges changed, not the distance.
By Coulomb's law, the force between two point charges q1 and q2 separated by a fixed distance d is
F=4πε01d2q1q2
Originally q1=+6 μC and q2=+9 μC, so F∝q1q2=54 (in μC2).
After −3 μC is added to each sphere:
q1′=6−3=3 μC, q2′=9−3=6 μC, so F′∝q1′q2′=18 (in μC2).
Since d is unchanged,
FF′=q1q2q1′q2′=5418=31
So F′=F/3.
✓Final answer(d) F/3.
- CBSE 2026Set ANNUAL1 markMCQQ.Two sphere of charge 2μc and 3μc are located at a distance 20 cm apart in air. The ratio of magnitude of electric forces acting between these spheres will be(a) 1 : 1(b) 2 : 3(c) 3 : 2(d) 4 : 9
›Reveal solutionSolution
The mutual electric force between two charges is an action-reaction pair, so both spheres feel equal magnitude forces regardless of the charge values.
By Coulomb's law the force sphere 1 exerts on sphere 2 has magnitude F = k q1 q2 / r^2, and the force sphere 2 exerts on sphere 1 has the same magnitude k q1 q2 / r^2, just opposite in direction (Newton's third law applies to electrostatic forces just as it does to mechanical ones). Since q1 q2 and r are common to both expressions, the two force magnitudes are identical no matter what q1 and q2 individually are.
✓Final answer(a) 1 : 1.
- CBSE 2025Set X11 markMCQQ.A point charge q1 exerts a force F on another point charge q2 when placed at a fixed distance. If another point charge q3 is brought near q2, the force on q2 due to q1 :(a) increases(b) decreases(c) may increase or decrease(d) does not change
›Reveal solutionSolution
(d) does not change. By the principle of superposition, the electrostatic force between q1 and q2 is given by Coulomb's law F=4πε01r2q1q2 and depends
✓Final answer(d) does not change.
By the principle of superposition, the electrostatic force between q1 and q2 is given by Coulomb's law F=4πε01r2q1q2 and depends only on q1, q2 and their separation. Bringing q3 near q2 adds a separate force on q2, but the force due to q1 is unaffected.
- CBSE 2025Set D1 markMCQQ.The distance between two charges is made half and one of the charges is also halved. The force acting between the two will become as compared to previous value (A) half (B) double (C) thrice (D) none of these
›Reveal solutionSolution
Coulomb force F ∝ q₁q₂/r²; halving one charge (×½) and halving the distance (×4) gives a net factor of 2, so the force doubles.
Coulomb's law:
F=r2kq1q2
Initial force: F=r2kq1q2.
Now one charge becomes q1/2 and the distance becomes r/2:
F′=(r/2)2k(q1/2)q2=r2/4kq1q2/2=24⋅r2kq1q2=2F
The force becomes double the previous value.
✓Final answer(B) double.
- CBSE 2025Set D1 markMCQQ.On inserting a dielectric material between two positive charges in air, the value of repulsive force will (A) increase (B) decrease (C) remain same (D) become zero
›Reveal solutionSolution
A dielectric weakens the field between charges by a factor K, so the repulsive force falls to F₀/K.
The Coulomb force between two charges in air is F₀ = (1/4πε₀)·q₁q₂/r². Filling the space with a dielectric of relative permittivity K replaces ε₀ with Kε₀:
F = (1/4πKε₀)·q₁q₂/r² = F₀/K
Since K > 1 for any dielectric, F < F₀. The force stays repulsive (both charges positive) but its magnitude decreases.
✓Final answer(B) decrease.
- CBSE 2025Set ANNUAL1 markMCQQ.The law governing the force between static electric charges is known as(i) Ampere's law(ii) Ohm's law(iii) Faraday's law(iv) Coulomb's law
›Reveal solutionSolution
The force between two static (point) electric charges is governed by Coulomb's law.
Ampere's law relates a magnetic field to the current producing it, Ohm's law relates current and voltage in a conductor, and Faraday's law deals with electromagnetic induction. None of these describes the force between charges at rest.
The electrostatic force between two point charges q1 and q2 separated by a distance r is F=4πε01r2q1q2 — this is Coulomb's law.
✓Final answer(iv) Coulomb's law.
- CBSE 2024Set IMPROVEMENT1 markMCQQ.On placing dielectric material between two point charges in air, repulsive force between them will —(a) Increase(b) Decrease(c) Remain same(d) Zero
›Reveal solutionSolution
Placing a dielectric between two charges reduces the force between them.
By Coulomb's law in a medium, F=4πε0K1r2q1q2, where K is the dielectric constant of the medium. Since a dielectric has K>1, placing it between the two point charges (in place of air/vacuum, K=1) reduces the force by a factor of K. This holds for both attractive and repulsive forces, so the repulsive force between the two positive charges will decrease.
✓Final answer(b) Decrease
- CBSE 2024Set FS1 markMCQQ.Force of 80 Newton works between two point charges placed at a fixed distance apart in air. When these charges are placed at the same distance apart in a dielectric medium, then force of 8 Newton works on it. The dielectric constant of medium will be:(i) K=−10(ii) K=10(iii) K=0.01(iv) K=−0.01
›Reveal solutionSolution
K=FmediumFair=880=10 — option (ii).
Concept. Coulomb's force between two charges at separation r is
Fair=4πε01r2q1q2,Fmedium=4πε0K1r2q1q2.
Placing a dielectric of constant K reduces the force by the factor K.
Solve. With the same charges and separation,
FmediumFair=K⇒K=880=10.
✓Final answer(ii) K=10
- CBSE 2024Set A1 markQ.Match Column 'A' with Column 'B' and write the correct pair. Column 'A' item: 'Electrostatic force'. Column 'B' options:(i) De-Broglie(ii) Maxwell(iii) Ohm(iv) Einstein(v) Coulomb(vi) Lenz(vii) Young.
›Reveal solutionSolution
Electrostatic force is governed by Coulomb's law.
The force of attraction or repulsion between two stationary point charges is called the electrostatic (or Coulomb) force, and its magnitude is given by Coulomb's law:
F=4πε01r2q1q2
This law was formulated by Charles-Augustin de Coulomb, so "Electrostatic force" pairs with option (v) Coulomb.
✓Final answerElectrostatic force → (v) Coulomb.
- CBSE 2024Set ANNUAL1 markMCQQ.Two charged spheres are separated by a distance d, exert a force F on each other. If the charges are doubled and the distance between them is doubled then the force is(a) F(b) F/2(c) F/4(d) 4F
›Reveal solutionSolution
Coulomb's law force scales as (charge product)/(distance)^2; doubling both charges and the distance leaves the force unchanged.
By Coulomb's law, the force between two point charges q1 and q2 separated by distance d is
F=4πϵ01d2q1q2=kd2q1q2
Now the charges are doubled (q1′=2q1, q2′=2q2) and the separation is doubled (d′=2d). The new force is
F′=k(2d)2(2q1)(2q2)=k4d24q1q2=kd2q1q2=F
So the force is exactly unchanged.
✓Final answer(a) F.
- CBSE 2024Set ANNUAL1 markQ.What is the name of the electrical force acting between two charges at rest?
›Reveal solutionSolution
The force between two charges at rest is the electrostatic (Coulomb) force.
The electrical force acting between two charges that are at rest (not moving) is called the electrostatic force, or Coulomb force, since it is governed by Coulomb's law:
F=4πϵ01r2q1q2
It acts along the line joining the two charges - repulsive for like charges, attractive for unlike charges. (This is distinct from the additional magnetic force that arises only when charges are in relative motion.)
✓Final answerThe Coulomb force (electrostatic force) between charges at rest.
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