Difficulty: Intermediate | Prerequisites: Electric charge, Coulomb's Law formula, vector addition of forces.
This material sits inside the electrostatics unit of a standard university physics sequence. The central question is deceptively simple: where can you place a charged particle so that all the electric forces on it cancel out? Answering it requires you to combine Coulomb's Law with vector reasoning, and that combination is the skill the rest of the course builds on (electric fields, Gauss's Law, electric potential). If you can find equilibrium positions confidently, you are in good shape for what comes next. You should already be comfortable with Coulomb's Law itself and the idea that like charges repel while opposite charges attract.
When two fixed charges sit on a line, there may (or may not) be a point where a test charge feels zero net force. Whether that point exists, and where it falls, depends on the signs and magnitudes of the charges and which side of them you look. Working through the algebra with Coulomb's Law and setting the two force magnitudes equal is the core technique.
Coulomb's Law
The force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them: F = k|q₁||q₂| / r². In simple terms, charges push or pull each other with a strength that drops off sharply as you move them apart.
Electrostatic equilibrium (for a test charge)
A position where the net electric force on a charge is zero because all individual forces cancel. Think of it as the "sweet spot" where the pushes and pulls from surrounding charges balance exactly.
Superposition principle
The net force on a charge is the vector sum of the forces from every other charge, each calculated independently. In simple terms, you just add up all the individual force arrows, and the total is what the charge actually feels.
Test charge
A charge placed in a region to probe the forces there. In these problems, the test charge is a proton (positive) or an electron (negative), and the fixed charges do not move.
Coulomb constant (k)
The proportionality constant in Coulomb's Law, approximately 8.99 × 10⁹ N·m²/C². It sets the scale for how strong electrostatic forces are in SI units.
When two positive charges sit on a line, a proton placed between them is repelled by both. The repulsion from the left charge pushes it right; the repulsion from the right charge pushes it left.
At the exact midpoint between two equal positive charges, the two repulsive forces are equal in magnitude and opposite in direction, so the net force is zero.
If the charges are not equal, the equilibrium point shifts towards the smaller charge. The test charge needs to be closer to the weaker source so that the weaker force, boosted by a shorter distance, can match the stronger source farther away.
Outside the pair (to the left of both, or to the right of both), both forces point in the same direction, so they cannot cancel. Equilibrium exists only between the two charges.
When one charge is positive and the other negative, the force landscape changes.
A proton between them is pushed away by the positive charge and pulled towards the negative charge. Both forces point in the same direction (away from the positive, towards the negative). They add rather than cancel, so no equilibrium exists between a positive and a negative charge.
An electron between them experiences the mirror image: attracted towards the positive charge and repelled by the negative charge, again both forces point the same way. No equilibrium between them either.
For an electron, the equilibrium point (if one exists) lies outside the pair, on the side of the charge whose magnitude is smaller, where the two forces can oppose each other.
Two positive charges, 4q at the origin and q at position x = L, both on the x-axis.
Part (a): proton to the right of q
Both 4q and q are positive, and the proton is positive. From the right side, both charges repel the proton to the right. The forces point in the same direction, so they cannot cancel. No equilibrium is possible to the right of q.
Part (b): proton between 4q and q
Place a proton at position r measured from 4q (so it is at distance L − r from q), with 0 < r < L.
F₄q points to the right (away from 4q), repelling the proton.
Fq points to the left (away from q), repelling the proton.
The two forces oppose each other. Because 4q is the larger charge, the proton must sit closer to q (the smaller charge) for the forces to balance. So yes, for the right choice of r, equilibrium is possible between the two charges.
Part (c): force expressions
Using Coulomb's Law with the proton's charge e:
F₄q = k(4q)(e) / r²
Fq = k(q)(e) / (L − r)²
Part (d): solving for the equilibrium position
Set F₄q = Fq:
k(4q)(e) / r² = k(q)(e) / (L − r)²
Cancel k, q, and e from both sides:
4 / r² = 1 / (L − r)²
Cross-multiply:
4(L − r)² = r²
Take the positive square root of both sides:
2(L − r) = r
2L − 2r = r
2L = 3r
r = 2L/3
The proton is in equilibrium two-thirds of the way from 4q to q. This is closer to q, which makes physical sense: the proton needs to be nearer to the weaker charge so that its force can compete with the stronger one farther away.
Coulomb's Law: F = k|q₁||q₂| / r²
Force on a proton (charge e) from charge 4q at distance r: F₄q = 4kqe / r²
Force on a proton from charge q at distance (L − r): Fq = kqe / (L − r)²
Equilibrium condition (net force = 0): 4kqe / r² = kqe / (L − r)² → r = 2L/3
Electrostatic equilibrium is not just a textbook exercise. Particle traps (used in mass spectrometry and quantum computing experiments) work by arranging electric fields so that charged particles sit at stable equilibrium points. The same logic, balancing forces from multiple sources, shows up whenever engineers position charged objects in electric fields, from ink droplets in inkjet printers to ions in accelerator beamlines.
Students often assume equilibrium can exist between a positive and a negative charge. It cannot, because both forces on a test charge point in the same direction there.
Students sometimes place the equilibrium point at the midpoint regardless of the charge magnitudes. The midpoint is correct only when the two charges are equal. Unequal charges shift the balance point towards the smaller one.
A common error is forgetting to take the positive square root when solving 4(L − r)² = r². Taking the negative root gives a position outside the two charges, which is physically invalid for this configuration (both forces would point the same way there).
Students occasionally set F₄q = Fq using r for both distances, rather than using r for one and (L − r) for the other. Always define your coordinate system clearly and express each distance in terms of a single variable.
⚠️ The technique of setting two Coulomb forces equal and solving for position comes up repeatedly in electrostatics exams. Practise it until the algebra is automatic.
⚠️ Being able to reason qualitatively about where equilibrium can and cannot exist (same-sign charges: between them; opposite-sign charges: outside, near the weaker one) saves time and catches algebraic mistakes.
⚠️ Problems may ask you to repeat this analysis with an electron instead of a proton. The equilibrium position does not change (the test charge's own charge cancels from both sides of the equation), but the stability of the equilibrium does.
⚠️ Watch the phrasing: "net force is zero" and "in equilibrium" mean the same thing here. Do not confuse this with stable equilibrium (where the particle returns after a small push), which is a separate question.
1. True or false: A proton can be in equilibrium between two positive charges of different magnitudes.
A: True. The equilibrium point just shifts towards the smaller charge.
2. Fill in the blank: For charges 4q and q separated by distance L, the equilibrium position for a test charge between them is at r = ______ from 4q.
A: 2L/3
3. True or false: If you replace the proton with an electron in the 4q and q problem, the equilibrium position changes.
A: False. The electron's charge cancels from both sides of the equation, so the position r = 2L/3 is the same.
4. True or false: A test charge can be in equilibrium between a positive charge and a negative charge.
A: False. Both forces point in the same direction in that region.
Q: Two positive point charges, Q and 9Q, are separated by distance d. Where between them is the net force on a proton zero? Express your answer as a distance from Q.
A: Set kQe/r² = k(9Q)e/(d − r)². Cancel common factors: 1/r² = 9/(d − r)². Take the positive square root: 1/r = 3/(d − r), so d − r = 3r, giving r = d/4 from Q. The equilibrium is one-quarter of the way from the smaller charge.
Q: Explain, without algebra, why no equilibrium point exists to the right of charge q (the smaller charge) when both charges are positive.
A: To the right of q, both 4q and q repel the proton in the same direction (to the right). Since neither force reverses direction, they cannot cancel, and the net force is always nonzero.
Q: Charges +3Q and −Q are placed on the x-axis at x = 0 and x = d. On which side of the arrangement could a proton be in equilibrium, and why?
A: To the right of −Q (the smaller-magnitude charge). Between the charges, both forces push/pull the proton in the same direction (towards −Q). To the left of +3Q, the stronger charge dominates at close range and still dominates at large range (since it is larger). To the right of −Q, the repulsion from +3Q (pointing right) and the attraction towards −Q (pointing left) oppose each other, and at the right distance they balance.
Q: In the 4q and q equilibrium problem, what happens to the equilibrium position if q is doubled to 2q (while 4q remains the same)?
A: The ratio of charges becomes 4q to 2q, or 2 to 1. Setting 2/(r²) = 1/(L − r)² gives √2(L − r) = r, so r = √2 L / (1 + √2) ≈ 0.586L. The equilibrium moves towards the midpoint because the charges are now closer in magnitude.
This connects directly to the concept of electric fields (Chapter 6 in most sequences), because the electric field is defined as force per unit charge. An equilibrium point for a test charge is the same as a point where the net electric field is zero, which becomes important when sketching field-line diagrams.
The algebraic technique here (setting two inverse-square expressions equal) reappears when finding the neutral point between gravitational sources, so if you have a mechanics background, the pattern is the same.
Understanding equilibrium stability (whether the particle returns to equilibrium or drifts away after a nudge) leads into Earnshaw's theorem, which proves that no static arrangement of charges can create a stable equilibrium in free space, a result with consequences for electrostatic trapping and plasma confinement.
Coulomb's Law, electrostatic equilibrium, net force zero, electric force balance, point charges, superposition of electric forces, Coulomb constant k, equilibrium position between charges, PHY 142, electricity and magnetism, inverse square law, test charge, force vector addition, charge ratio equilibrium, two-charge equilibrium problem