Source: University Physics, Electricity & Magnetism – Discussion Activity 1
Tags: Coulomb's law, electrostatic equilibrium, net force zero, point charges, superposition of forces, electric force balance, PHY 142, electricity and magnetism
Difficulty: Intermediate | Prerequisites: Basic vector addition, Newton's third law, algebraic manipulation of fractions.
Big picture: This material sits at the heart of the electrostatics unit. Coulomb's law tells you the force between two point charges; this chapter asks what happens when a third charge sits between (or beside) two others and you need the net force on it to vanish. That zero-net-force condition, electrostatic equilibrium, is one of the most common exam setups in introductory E&M. You need to be comfortable with Coulomb's law itself, the principle of superposition, and basic algebra to follow along.
When two fixed point charges sit on an axis, there is exactly one location where a third charge feels zero net force. For like charges, that point is always between them (closer to the smaller charge). For opposite charges, the equilibrium point is outside, beyond the smaller-magnitude charge. Finding it is an algebra problem: set two Coulomb forces equal and solve for position.
Coulomb's law
The electrostatic 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, bigger charges or shorter distances mean stronger forces.
Electrostatic equilibrium (zero net force)
A condition where the vector sum of all electric forces on a charge is zero, so the charge would remain stationary if released. Think of it as the electrical version of a balanced see-saw: the pulls and pushes cancel exactly.
Superposition principle
The net electric force on a charge is the vector sum of the individual forces from every other charge present. In simple terms, you calculate each force separately, then add them as vectors.
Coulomb constant (k)
The proportionality constant in Coulomb's law, k ≈ 8.99 × 10⁹ N·m²/C². It sets the scale of how strong electrostatic forces are in SI units.
Test charge
A charge (often a proton or electron) placed in a region to probe the electric environment. It is assumed small enough not to disturb the source charges.
When two positive charges sit on a line, a proton placed between them can experience zero net force. The repulsion from the left charge pushes it right; the repulsion from the right charge pushes it left. At exactly the right spot, those pushes balance.
If the two charges are equal, the equilibrium point is the midpoint.
If the charges are unequal (say 4q and q), the equilibrium point shifts toward the smaller charge. The larger charge's force drops off with distance squared, so you need to be further from it and closer to the smaller one to get a balance.
An electron between two positive charges also finds equilibrium at the same location. The direction of each force flips (attraction instead of repulsion), but both flip, so the balance point does not move. The equilibrium position depends on the magnitudes of the source charges, not the sign or size of the test charge.
If a proton is placed to the right of both positive charges, both Coulomb forces point to the right (both repel). The forces are in the same direction, so they cannot cancel. The same logic applies to the left of both charges.
For a proton between charges 4q and q: to the right of q, both 4q and q push the proton further right, so no equilibrium exists there.
Place 4q at the origin and q at position x = L. A proton (charge e) sits at position x = r, where 0 < r < L (between the two charges).
Distance from 4q to the proton: r
Distance from q to the proton: (L − r)
The two forces on the proton:
F₄q (from the 4q charge, pushing the proton to the right):
$$F_{4q} = \frac{k(4q)(e)}{r^2}$$
Fq (from the q charge, pushing the proton to the left):
$$F_q = \frac{k(q)(e)}{(L - r)^2}$$
Set F₄q = Fq (magnitudes equal, directions opposite):
$$\frac{k(4q)(e)}{r^2} = \frac{k(q)(e)}{(L - r)^2}$$
Cancel k, q, and e from both sides:
$$\frac{4}{r^2} = \frac{1}{(L - r)^2}$$
Take the square root of both sides:
$$\frac{2}{r} = \frac{1}{L - r}$$
Cross-multiply:
$$2(L - r) = r$$
$$2L - 2r = r$$
$$2L = 3r$$
$$r = \frac{2L}{3}$$
The proton is in equilibrium two-thirds of the way from 4q toward q, which is the same as one-third of L from q. This makes physical sense: the proton must sit closer to the weaker charge for the forces to balance.
Quantity | Expression |
|---|---|
Coulomb's law | F = k|q₁||q₂| / r² |
Force from 4q on proton | F₄q = k(4q)(e) / r² |
Force from q on proton | Fq = k(q)(e) / (L − r)² |
Equilibrium condition | 4 / r² = 1 / (L − r)² |
Equilibrium position | r = 2L/3 (measured from 4q) |
Electrostatic equilibrium is the principle behind electrostatic precipitators in power plants, where charged particles are guided to collection plates by carefully balanced electric fields. It also underpins how charged toner particles are positioned precisely in laser printers: the forces from multiple charged regions must balance so the toner lands in the right spot.
"The equilibrium point is always at the midpoint." Only true when the two source charges have equal magnitude. With unequal charges, the point shifts toward the smaller one.
"A negative test charge would have a different equilibrium position." The position depends on where force magnitudes balance, and flipping the test charge sign flips both forces simultaneously. The equilibrium location is the same for a proton or an electron (though stability differs).
"If both charges are positive, there could be equilibrium outside both charges." There cannot. Outside both charges, both forces push in the same direction. Equilibrium between like charges is always between them.
"You need to know the value of the test charge to find the equilibrium position." The test charge cancels out of both sides of the equation. The position depends only on the ratio of the source charges and the separation L.
⚠️ Setting up the distance expressions correctly is where most errors occur. If 4q is at the origin and q is at L, and the proton is at position r from 4q, then its distance to q is (L − r), not (r − L) or just L.
⚠️ Forgetting to take the square root before solving is a common algebra slip. You can also cross-multiply directly from 4(L − r)² = r², but the square-root route is cleaner.
⚠️ The equilibrium position r = 2L/3 is measured from 4q. Exams sometimes ask for the distance from q, which is L/3. Read the question carefully.
⚠️ Qualitative reasoning (can equilibrium exist in this region?) is tested as often as the algebra. Be ready to explain, in one or two sentences, why forces in a given region do or do not oppose each other.
True or false: A proton placed to the right of both positive charges 4q and q can be in electrostatic equilibrium.
A: False. Both forces push it further to the right.
Fill in the blank: For two positive charges Q and q (Q > q) separated by distance L, the equilibrium point is closer to ______.
A: The smaller charge, q.
True or false: The equilibrium position changes if you replace the proton with an electron.
A: False. The test charge cancels from the equilibrium equation.
Fill in the blank: If 4q is at the origin and q is at x = L, the equilibrium position is at x = ______.
A: 2L/3.
Q: Two point charges, +9Q and +Q, are separated by distance d. Where on the line between them is the net electric force on a proton equal to zero? Express your answer as a distance from +9Q.
A: Using the same method, 9/(r²) = 1/(d − r)², giving 3/r = 1/(d − r), so 3(d − r) = r, hence r = 3d/4. The equilibrium point is three-quarters of the way from +9Q toward +Q.
Q: Charges +4q and +q are separated by distance L. A proton is placed on the x-axis to the right of +q. Is equilibrium possible? Explain in one sentence.
A: No, because both charges repel the proton in the same direction (to the right), so the forces cannot cancel.
Q: Write the equilibrium condition for a test charge between +4q (at origin) and +q (at x = L), then solve for the position.
A: Set k(4q)(e)/r² = k(q)(e)/(L − r)². Cancel common factors, take the square root: 2/r = 1/(L − r). Solve to get r = 2L/3.
Q: If the charges were +4q and −q instead (one positive, one negative), in which region(s) could a proton be in equilibrium?
A: The equilibrium point would be outside the pair, beyond the −q charge (to the right of −q). Between them, both the attraction from −q and the repulsion from +4q push the proton in the same direction. Beyond 4q on the left, the larger charge dominates. Only beyond −q can the forces oppose each other and balance.
This material connects directly to electric fields (Chapter 6 in most texts): the equilibrium point for a test charge is also the point where the net electric field is zero, since F = qE. Understanding force balance here prepares you for field-line diagrams and equipotential surfaces.
It also links to gravitational analogues in mechanics. The Lagrange points between two massive bodies (say the Earth and the Moon) are found by the same style of calculation, balancing inverse-square forces. If you have seen L1 points in an astronomy or orbital mechanics context, the algebra is nearly identical.
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