Q.A cubic unit cell of cesium chloride (CsCl) has cesium atoms situated at the eight corners of a cube of side 0.40 nm, and a chlorine atom situated at the centre of the cube. Each Cs atom is deficient by one electron, so it carries a charge +e; the Cl atom carries one excess electron, so it carries a charge −e.
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🔒 Start your 14-day free trial to unlock the full solution →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.
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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.
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The eight corner charges are arranged symmetrically about the centre, so their fields cancel: the net field at Cl is zero. Removing one corner charge leaves a net force equal and opposite to what that one charge contributed — magnitude ≈1.9×10−9N, directed away from the vacant corner A.
(i) Net field =0 by symmetry. …
Because the eight Cs+ charges sit at corners that pair up diametrically across the centre, their fields at the Cl atom cancel and the net field is zero. When one Cs+ is removed, the remaining seven give a net force equal in magnitude and opposite in direction to the force the removed charge alone would have exerted — about 1.92×10−9N, pointing away from the missing corner.
(i) Net field at Cl
The eight corners occur in four pairs, each pair lying on a body diagonal on opposite sides of the centre and equidistant from it. The two Cs+ of a pair produce equal and opposite fields at the centre, which cancel. Summing over the four pairs:
Enet=0.
(ii) Net force after removing the corner charge at A
Use superposition. With all eight present the force on Cl is zero:
F8=FA+Fother 7=0⇒Fother 7=−FA.
So the seven remaining charges give a force equal and opposite to the force FA that the removed Cs+ at A exerted on Cl. Since Cs+ attracted the Cl− toward A, −FA points away from A (along the body diagonal, toward the diagonally opposite corner). Equivalently, removing a +e is like adding a −e at A, which repels the Cl−.
Magnitude. The corner-to-centre distance is half the body diagonal: …
Method: Using Symmetry to Simplify Superposition, Then Removing One Term
Use this whenever many identical charges surround a central point symmetrically, and the question then asks what happens when ONE of them is removed or changed.
Steps
Step 1: Check for pairwise symmetry in the FULL arrangement first.
If equal-magnitude, equal-sign charges sit at positions that pair up symmetrically about the point of interest (e.g. opposite corners of a cube through its centre), each pair's field/force contributions cancel exactly. Conclude the full arrangement gives zero net field/force before doing any further algebra.
Step 2: When one charge is removed, use superposition algebraically instead of resumming from scratch.
Fall=Fremoved charge alone+Fremaining charges
Since Fall=0 from Step 1, it follows that Fremaining=−Fremoved charge alone — exactly equal and opposite to whichever single charge's contribution was taken out. …
- GUJCET 2026Set x1 markMCQQ.Two infinitely long thin straight parallel wires are kept a perpendicular distance 2R having uniform linear charge densities +λ and −λ respectively. The magnitude of electric field at a mid point between two wires will be ______. (A) πε0Rλ (B) 2πε0Rλ (C) πε0R2λ (D) 4πε0Rλ
›Reveal solutionSolution
Fields of the +λ and −λ wires add at the midpoint: E=λ/πε0R.
Field of an infinite line at distance r: E=2πε0rλ. The midpoint is at r=R from each wire. …
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.The electrostatic force on a small sphere of charge 0.4μC due to another small sphere of charge -0.8μC in air is 0.2 N. What is the distance between the two spheres?(a) 12 m(b) 0.12 m(c) 1.2 m(d) 0.012 m
›Reveal solutionSolution
Coulomb's law relates the electrostatic force between two point charges to their separation: F = kq1q2/r².
Given q1 = 0.4 μC = 4 × 10⁻⁷ C, q2 = 0.8 μC = 8 × 10⁻⁷ C (magnitudes), F = 0.2 N, k = 9 × 10⁹ N m²/C².
…
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.Two identical conducting spheres A and B having charges +q and -q are kept at 'd' distance apart experience coulombian force F between them. If 50% of charge is transferred from sphere B to A then the new coulombian force between them is ___.(a) F(b) F/2(c) F/4(d) 2F/3
›Reveal solutionSolution
Coulomb's force is proportional to the product of the two charges; recompute the new charges after the transfer and rescale F accordingly.
Original force: F = k q (q)/d² (magnitude, using |+q| and |−q| = q each).
50% of sphere B's charge (−q) is transferred to A: transferred charge = −q/2. …
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.As shown in figure charges +q each are placed at the four vertices of a square. Then the coulombian force acting on charge placed at vertex D is ___.(a) (√2 + 1/2) kq^2/a^2(b) (√2 - 1/2) kq^2/a^2(c) √2 kq^2/a^2(d) kq^2/2a^2
›Reveal solutionSolution
The net force on a corner charge in a square of equal charges is the vector sum of two equal edge forces (perpendicular to each other) and one diagonal force.
Charge at D experiences:
- Force from A (distance a, along DA): magnitude kq²/a²
- Force from C (distance a, along DC): magnitude kq²/a², perpendicular to the A-force
- Force from B (diagonal, distance a√2): magnitude kq²/(a√2)² = kq²/(2a²), directed along the diagonal DB …
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.The Coulombian repulsive force between two alpha particles kept at a distance of 3 cm in air is ___ N.(a) 1.024 x 10^-27(b) 1.024 x 10^-25(c) 1.024 x 10^-24(d) 1.024 x 10^-23
›Reveal solutionSolution
Alpha charge = 2e = 3.2x10^-19 C; Coulomb's law with r = 0.03 m gives F = 1.024 x 10^-24 N.
Each alpha particle has charge q = 2e = 3.2 x 10^-19 C. Separation r = 3 cm = 0.03 m.
Coulomb force: F = k q^2 / r^2 = (9 x 10^9)(3.2 x 10^-19)^2/(0.03)^2.
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- GUJCET 2021Set 151 markMCQQ.Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of same signs and of magnitude 17.7×10−22 C/m2. What is E in the outer region of the second plate? (A) 4×10−10 NC−1 (B) 2×10−10 NC−1 (C) 1×10−10 NC−1 (D) Zero
›Reveal solutionSolution
Same-sign charged plates give E=σ/ε0 in the outer region (fields add).
Concept: Each sheet produces 2ε0σ. In the region outside the second plate both fields point the same way and add: …
- GUJCET 2020Set 071 markMCQQ.Two point electric charges +10−8 C and −10−8 C are placed 0.1 m apart. Find the magnitude of Total Electric Field at the center of the line joining the two charges. (A) Zero (B) 3.6×104NC−1 (C) 7.2×104NC−1 (D) 12.96×104NC−1
›Reveal solutionSolution
At the centre of a dipole-like pair, both fields point the same way and add.
Concept — superposition of fields. At the midpoint, the field of +q points away from it and the field of −q points toward it — both in the same direction, so they add.
Steps.
- Distance from each charge: r=0.05 m. …
- GUJCET 2019Set 131 markMCQQ.When two sppheres having 4Q and −2Q charge are placed at a certain distance, the force acting between them is F. Now they are connected by a conducting wire and again separated from each other. Now they are kept at a distance half of the previous one. The force acting between them is .......... (A) 8F (B) 2F (C) 4F (D) F
›Reveal solutionSolution
Charges redistribute to Q each, and at half the separation the force is F/2.
Concept: When two conductors are joined by a wire the total charge shares equally. Coulomb force F=r2kq1q2.
Steps:
- Initial magnitude: F=d2k(4Q)(2Q)=d28kQ2.
- After connection each sphere has 24Q+(−2Q)=Q. …
- GUJCET 2019Set 131 markMCQQ.Charge of 1μC each is placed on the five corners of a ragular hexagon of side 1m. The electric field at its centre is ...........N/C. (A) 10−6K (B) 56×10−6K (C) 5×10−6K (D) 65×10−6K
›Reveal solutionSolution
Missing one of six symmetric charges leaves a net field equal to a single charge's field, 10−6K.
Concept: By symmetry, six equal charges at the vertices of a regular hexagon produce zero field at the centre (each field cancels its diametric opposite). Removing one charge is equivalent to superposing the full symmetric set (field 0) with a single negative-of-that charge at that vertex, leaving the field of one charge.
Steps: …
- GUJCET 2015Set C1 markMCQQ.A point charge q is situated at a distance r on axis from one end of a thin conducting rod of length L having a charge Q [Uniformly distributed along its length]. The magnitude of electric force between the two is _____. (A) r2KQq (B) r(r+L)2KQ (C) r(r−L)KQq (D) r(r+L)KQq
›Reveal solutionSolution
[!TLDR] Integrating the point-charge force over the uniformly charged rod gives F=r(r+L)KQq.
Concept
A charge distributed along a line is handled by integration: split it into elements dq, write the Coulomb force dF=x2Kqdq from each element at distance x, and integrate. Here all forces are collinear (rod on the axis), so they add as scalars.
Solution
Linear charge density λ=LQ, so dq=LQdx. The near end of the rod is at distance r, the far end at r+L. …
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