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Two charges are fixed on a line. Where can you put a third charge so that the net force on it is zero? And what charge must it carry so that all three are in equilibrium? These questions use nothing beyond Coulomb's law and superposition, yet sign and position slips are common. This post gives a decision rule, two worked examples with checks, and a note on stability.
Is this in the syllabus?
NCERT's chapter Electric Charges and Fields covers Coulomb's law in section 1.5 and forces between multiple charges, using the superposition principle, in section 1.6. The CBSE Physics curriculum for 2026-27 and the Syllabus for JEE (Main) 2026 both list "forces between multiple charges" and the superposition principle, and the JEE (Advanced) 2026 Information Brochure lists "Coulomb's law". Check the current year's documents before your exam.
Where can the third charge sit?
Let q₁ and q₂ be fixed a distance d apart. A third charge q feels two forces along the line. For them to cancel, they must point in opposite directions and be equal in size.
Like charges (both positive or both negative): the point lies between them, closer to the smaller charge.
Unlike charges: the point lies outside, beyond the charge of smaller magnitude.
Equal and opposite charges: there is no such point at a finite distance.
The sign and size of q do not matter for its own equilibrium, because q multiplies both forces. Write kqq₁/r₁² = kqq₂/r₂², cancel q, and take square roots.
Worked example 1: between two like charges
Charges of +4 μC and +9 μC are fixed 0.50 m apart. Find where a third charge feels no force, and the charge it must carry for all three to be in equilibrium.
Let the point be x from the +4 μC charge. Then 4/x² = 9/(0.50 − x)².
Square roots: 2/x = 3/(0.50 − x), so 2(0.50 − x) = 3x and x = 0.20 m.
For the +9 μC charge to be in equilibrium too, the third charge must pull it with a force equal to the push from +4 μC. So it is negative, with k|q| × 9/0.30² = k × 4 × 9/0.50², giving |q| = 4 × 0.09/0.25 = 1.44 μC. The third charge is −1.44 μC.
Check the +4 μC charge: the push from +9 μC is k × 4 × 9/0.25 = 144k; the pull from −1.44 μC is k × 4 × 1.44/0.04 = 144k. Balanced.
Size of each force: with k = 1/(4πε₀) ≈ 8.99 × 10⁹ N m²/C² (from the NIST value of ε₀), 144 × 10⁻¹² C²/m² × k ≈ 1.29 N.
A shortcut for step 3: the third charge is −q₂(x/d)² = −9 × (0.20/0.50)² = −1.44 μC.
Worked example 2: outside two unlike charges
Charges of +4 μC and −1 μC are fixed 0.30 m apart. Where does a third charge feel no force?
The smaller charge is −1 μC, so the point lies beyond it, at a distance r from it.
4/(0.30 + r)² = 1/r², so 2r = 0.30 + r and r = 0.30 m.
The point is 0.30 m beyond the −1 μC charge, 0.60 m from the +4 μC charge.
Between the charges, both forces on a test charge point the same way, so no balance is possible there. Beyond the larger charge, the larger charge is also the closer one, so its force always wins.
Is the equilibrium stable?
Take example 1 with a positive test charge at the zero-force point. Nudge it along the line toward +9 μC: the push from +9 μC grows and the push from +4 μC weakens, so it is driven back. Along the line, it is stable. Nudge it sideways: both pushes now have components away from the line, so it moves further off. Sideways, it is unstable. A negative test charge does the opposite: unstable along the line, stable sideways. Questions often ask about one direction only, so read which displacement they mean.
Which mistakes cost marks?
Placing the point between two unlike charges.
Putting the point nearer the larger charge.
Keeping the wrong root. Squaring gives two solutions; keep the one in the correct region.
Giving the third charge the wrong sign when all three must be in equilibrium. For two like charges, it must be opposite.
Forgetting to convert μC and cm to SI units when a force is asked for.
Why does electrostatics feel harder than it looks?
The Conceptual Survey of Electricity and Magnetism, described by Maloney and colleagues (2001), was given in some form to more than 5000 introductory physics students. Typical scores after instruction were only 47% in calculus-based courses and 44% in algebra-based ones. Short, reasoning-first problems like these show where the gaps are. Our profile of Charles-Augustin de Coulomb tells the story behind the law.
A practice routine
Day 1: for six charge pairs, sketch the line and mark the region where the zero-force point must lie, before any algebra.
Day 2: solve three of them, with the check line from example 1.
Day 3: two all-three-in-equilibrium problems and one stability question.
Day 4: a mixed set that adds a problem from our post on two charged spheres joined by a wire.
When the force arrows will not balance, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Tonight, redo example 1 with +1 μC and +4 μC placed 0.60 m apart, and check that the zero-force point is 0.20 m from the smaller charge.
