Q.The electrostatic force on a small sphere of charge 0.4μC due to another small sphere of charge −0.8μC in air is 0.2N.
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: Inverse Square Law Comparison — Coulomb’s law gives the force between two point charges as F=kr2∣q1q2∣, where k=9×109 N m2/C2 in air.
(a)
F=0.2 N, q1=0.4×10−6 C, q2=−0.8×10−6 C.
Using r2=kF∣q1q2∣:
r2=9×109×0.2(0.4×10−6)(0.8×10−6)=9×109×0.23.2×10−13=9×109×1.6×10−12=1.44×10−2.
So r=1.44×10−2=0.12 m.
(b)
By Newton’s third law, the force on the second sphere due to the first is equal in magnitude and opposite in direction: 0.2 N, attractive.
The distance is 0.12 m and the force on the second sphere is 0.2 N (attractive).
Using Coulomb’s law, the distance is found from F=kr2∣q1q2∣, and by Newton’s third law the force on the second sphere is equal in magnitude and opposite in direction to the force on the first. The distance is 0.12m and the force on the second sphere is 0.2N (attractive).
The problem is a direct application of Coulomb’s law for the electrostatic force between two point charges. The key idea is that the force magnitude depends only on the product of the charges and the square of the distance between them — the sign of the charges tells us the direction (attractive or repulsive), but the magnitude is given by the absolute values.
Because the two charges are opposite in sign, the force is attractive. The problem gives the force on the first sphere, and part (b) simply asks for the force on the second sphere — which, by Newton’s third law, must be equal in magnitude and opposite in direction.
Let’s work through it step by step.
- Write down Coulomb’s law in magnitude form The electrostatic force between two point charges q1 and q2 separated by a distance r in vacuum (or air, which has nearly the same permittivity) is:
F=kr2∣q1q2∣
where k=4πε01=9×109N m2/C2.
-
Identify the given quantities
- q1=0.4μC=0.4×10−6C=4×10−7C
- q2=−0.8μC=−0.8×10−6C=−8×10−7C
- F=0.2N (magnitude of force on q1 due to q2)
The product ∣q1q2∣=(4×10−7)(8×10−7)=32×10−14=3.2×10−13C2.
-
Solve for the distance r
Rearranging Coulomb’s law:
r2=kF∣q1q2∣
Substitute the values:
r2=(9×109)×0.23.2×10−13
First compute the fraction:
0.23.2×10−13=1.6×10−12
Then:
r2=9×109×1.6×10−12=14.4×10−3=1.44×10−2
Taking the square root:
r=1.44×10−2=1.44×10−1=1.2×10−1=0.12m
So the distance between the spheres is 0.12 metres (or 12 cm).
Notice that we used the magnitude of the charges. The negative sign on q2 only tells us the force is attractive — it doesn’t affect the distance calculation.
- Answer part (b) using Newton’s third law The force on the second sphere due to the first is equal in magnitude and opposite in direction to the force on the first sphere due to the second. Magnitude: 0.2N Direction: Since the charges are opposite, the force is attractive — so the second sphere is pulled toward the first. Thus the force on the second sphere is 0.2N (attractive).
A common mistake is to think the force on the second sphere is different because the charges have different magnitudes. But Coulomb’s law gives the force on each charge as the same magnitude — the product ∣q1q2∣ is symmetric. Newton’s third law guarantees equality.
The distance between the spheres is 0.12m and the force on the second sphere due to the first is 0.2N (attractive).
Method: Coulomb’s Law (Inverse Square Law)
We use Coulomb’s Law for electrostatic force between two point charges:
F=kr2∣q1q2∣
where:
- F = magnitude of electrostatic force (N)
- k=9×109 N m2/C2 (Coulomb’s constant for air)
- q1,q2 = charges (C)
- r = distance between charges (m)
Step 1: Identify given values
- q1=0.4 μC=0.4×10−6 C
- q2=−0.8 μC=−0.8×10−6 C
- F=0.2 N
Note: Force magnitude uses absolute values of charges. The negative sign on q2 only tells us the force is attractive.
Step 2: Solve for distance r (part a)
From Coulomb’s Law:
r2=kF∣q1q2∣
Substitute values:
r2=(9×109)×0.2(0.4×10−6)×(0.8×10−6)
Simplify numerator:
0.4×0.8=0.32
10−6×10−6=10−12
So ∣q1q2∣=0.32×10−12 C2
Now:
r2=9×109×0.20.32×10−12
r2=9×109×1.6×10−12
r2=14.4×10−3=0.0144
Take square root:
r=0.0144=0.12 m
Answer (a): 0.12 m (or 12 cm)
Step 3: Force on second sphere (part b)
By Newton’s Third Law, the force on the second sphere due to the first is equal in magnitude and opposite in direction to the force on the first sphere due to the second.
- Magnitude: 0.2 N
- Direction: attractive (toward the first sphere)
Answer (b): 0.2 N (attractive, toward the first sphere)
Key Concept Check
- The inverse square law means: if distance doubles, force becomes 41.
- Force magnitude depends only on product of charges and distance — not on which charge we consider.
- The negative sign on q2 indicates opposite charges → attraction.
Common Mistakes & How to Avoid Them
Mistake 1: Forgetting to convert microcoulombs (μC) to coulombs (C)
The error: Students plug 0.4 and 0.8 directly into Coulomb's law, forgetting the μ (micro) means 10−6.
Why it happens: The problem gives charges in μC, but Coulomb's law requires SI units (C). The conversion factor is easy to overlook under time pressure.
How to avoid: Always write the conversion explicitly before substituting:
q1=0.4μC=0.4×10−6C=4×10−7C
q2=−0.8μC=−0.8×10−6C=−8×10−7C
Pro tip: Circle or underline the unit in the question. If it's not in base SI, convert first — every time.
Mistake 2: Using the wrong value of k or forgetting it entirely
The error: Some students use k=9×109 but forget the units, or mistakenly use k=1 (thinking "in air" means vacuum permittivity is irrelevant).
Why it happens: The constant k=4πε01=9×109N m2/C2 is often memorised without understanding its role.
How to avoid: Write Coulomb's law fully:
F=kr2∣q1q2∣
Then substitute k=9×109 with its units. This helps you check that your final distance comes out in metres.
Mistake 3: Ignoring the sign of the charges when calculating force magnitude
The error: Students include the negative sign of q2=−0.8μC in the product q1q2, getting a negative value, then panic or get confused.
Why it happens: Coulomb's law for magnitude uses absolute values. The sign only tells you attraction (opposite signs) or repulsion (same sign).
How to avoid: For part (a), use only magnitudes:
F=kr2∣q1∣⋅∣q2∣
So ∣q1∣=4×10−7C and ∣q2∣=8×10−7C. The force 0.2N is already positive — it's the magnitude.
Key insight: The sign of the force (attractive/repulsive) is a direction concept, not a magnitude concept. Part (a) only asks for distance, so signs are irrelevant.
Mistake 4: Solving for r incorrectly (algebra errors)
The error: After substituting, students make mistakes like:
- Forgetting to take the square root
- Inverting the fraction
- Misplacing powers of 10
How to avoid: Solve step-by-step:
- Write: r2=kF∣q1q2∣
- Substitute carefully:
r2=(9×109)×0.2(4×10−7)(8×10−7)
- Simplify powers of 10 separately:
=9×109×0.232×10−14
=9×109×160×10−14
=1440×10−5=1.44×10−2
- Take square root: r=1.44×10−2=1.2×10−1=0.12m
Check: 0.12m=12cm — a reasonable distance for these charges and force.
Mistake 5: Answering part (b) with a different value than 0.2N
The error: Students recalculate the force using the distance found in (a), but make an arithmetic slip, or think the force on the second sphere is somehow different.
Why it happens: They forget Newton's Third Law — electrostatic forces are action-reaction pairs.
How to avoid: Remember: The force on sphere 2 due to sphere 1 is equal in magnitude and opposite in direction to the force on sphere 1 due to sphere 2.
So part (b) answer is simply:
0.2N
Direction: Attractive (since charges are opposite), but the question only asks for force magnitude.
Mistake 6: Giving the distance in wrong units
The error: After calculating r=0.12m, students write the answer as 0.12 without units, or convert unnecessarily to cm without being asked.
How to avoid: Always state the unit. The standard SI unit for distance is metres. Write:
0.12m
If you prefer, you can add (12cm) in brackets, but the primary answer should be in metres unless the question specifies otherwise.
Quick Summary Checklist
| Mistake | How to Avoid |
|---|---|
| Forgetting μ conversion | Convert to C before plugging in |
| Wrong k value | Write k=9×109 with units |
| Including sign in magnitude | Use $ |
| Algebra errors in r | Solve stepwise, check powers of 10 |
| Wrong force in (b) | Newton's Third Law: same magnitude |
| Missing units | Always write the unit with the number |
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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