Electric charge and Coulomb’s law

One charge, two different directions

Imagine three very small charged objects held at marked positions. Object C is positive. Positive A pushes C away, while negative B pulls C toward it. The two forces point in different directions. Where does their combined force point?

We will answer with one concrete arrangement: A is at (−3 cm, 0), B at (3 cm, 0), and C at (0, 4 cm). Their assigned charges are +4 nC, −1 nC and +1 nC. First predict the directions; then calculate the components. These are ideal static point charges in vacuum, not a suggested experiment with charged equipment.

The supports keep the positions fixed. A nonzero electric force on C does not mean that C is already moving: a support can supply an opposing force. A force arrow is not a trajectory.

Charge is a signed scalar

Electric charge is a property involved in electromagnetic interactions. Its sign is positive or negative; an uncharged or electrically neutral object has zero net charge. Positive and negative are labels for the two signs, not labels for motion, mass or goodness.

Add charge amounts algebraically. Equal-magnitude positive and negative amounts have a zero scalar sum. There is no arrow direction to cancel: charge is a scalar, while force is a vector. Rotating the coordinate axes changes force components but not an object’s charge.

Net neutrality does not mean that charged constituents are absent. Ordinary neutral matter contains positive and negative charges whose total is zero. Their spatial arrangement can still matter to another charged object.

Name the unit before using a number

The SI unit of charge is the coulomb, C. The unit relation is 1 C = 1 A·s, where A is the ampere and s the second. Charge is an amount; current describes a rate of charge flow. A charge of one coulomb is not a current of one ampere unless a specified transfer takes one second.

The elementary charge is exactly e = 1.602176634 × 10⁻¹⁹ C in the present SI. An electron has charge −e and a proton +e. For electron transfer between ordinary objects, a whole number N of electrons changes their charges by opposite amounts of magnitude Ne.

Our prefix n means nano: 1 nC = 10⁻⁹ C. The assigned nanocoulomb values in the static model are classical idealizations, not assertions that these exact amounts were obtained by counting a whole number of electrons. Do not infer exact particle counts by treating rounded macroscopic charges as exact multiples of e.

Quantization statements need their setting. Quarks have fractional charges but are not freely isolated ordinary charged objects. Fractionally charged quasiparticles also occur in particular many-body states, such as the fractional quantum Hall regime. That does not mean that an isolated electron has been divided into three freely available pieces.

Transfer is different from polarization

If electrons move from one initially neutral object to another, the first becomes positive and the second negative. For an isolated combined system, the gain of negative charge on one is balanced by its loss on the other. Charging by contact or rubbing does not create net electric charge from nothing.

A different response is redistribution within an object. In a conductor, mobile charges can rearrange; in an insulator, positive and negative constituents can shift relative to one another. This polarization can leave the whole object neutral.

A neutral polarizable object can be attracted toward a region of stronger electric field. In a nonuniform field, the forces on its separated charge regions need not cancel. Consequently, attraction alone is not a reliable test that two whole objects carry opposite net charges. A full material response depends on the object, its surroundings and the field geometry.

State the static point-charge law

For two distinct, fixed point charges in vacuum, the magnitude of their mutual electric force is F = kₑ |q₁q₂|/r². Here r > 0 is their separation, |q₁q₂| is the nonnegative magnitude of the charge product, and kₑ = 1/(4π ε₀), with ε₀ the vacuum electric permittivity.

The force lies along the line joining the points. Equal signs give repulsion; opposite signs give attraction. Decide which object receives the force before drawing its arrow. “The charges attract” describes a pair interaction, not one arrow that belongs to both objects.

The SI units of kₑ are N·m²/C², so the magnitude formula gives newtons. The exact SI definition of e does not make the numerical SI value of ε₀ or kₑ exact: these electromagnetic constants are determined experimentally. We keep kₑ symbolic in the example so no rounded constant is silently treated as a definition.

Use the source-to-recipient displacement once

Let qⱼ be a source charge at position rⱼ and qᵢ the recipient at rᵢ. Define dᵢⱼ = rᵢ − rⱼ, pointing from source j to recipient i. The vector law is Fᵢ←ⱼ = kₑ qᵢqⱼ dᵢⱼ/|dᵢⱼ|³.

The vector in the numerator has length |dᵢⱼ|; dividing by its cube therefore leaves the inverse-square magnitude. If qᵢqⱼ is positive, the force follows dᵢⱼ and pushes i away. If the product is negative, the force reverses dᵢⱼ and pulls i toward j.

There are two consistent methods: calculate a nonnegative magnitude and assign its direction, or use the signed vector formula. Do not add another attraction minus sign after a negative charge product has already reversed the vector. Exclude a charge’s own contribution; dᵢᵢ = 0 is not an allowed term.

Change distance with the other inputs fixed

Keep the same two assigned charges and compare the force at separation r with its magnitude F(r₀) at a positive reference separation r₀. Their ratio is F(r)/F(r₀) = (r₀/r)². Doubling the distance reduces the magnitude to one quarter, not one half.

Inverse-square scaling describes this fixed-charge point model. It is not a promise that a finite conductor keeps the same charge distribution as a neighboring object approaches. Nor does the divergence of the formula as r approaches zero establish a physically measurable infinite force: finite size, microscopic structure and a different physical theory can become essential.

The comparison uses positive separations only. When reading a graph, check the actual axes and endpoint domain; a curve must not disguise r = 0 or replace a missing singular point with an arbitrary finite value.

Choose units for the three-charge case

Use q₀ = 1 nC and L₀ = 1 cm. Express positions as dimensionless vectors R = r/L₀ and forces in the unit F₀ = kₑ q₀²/L₀². Thus F₀ = kₑ × 10⁻¹⁴ C²/m²; multiplying by kₑ’s units gives newtons. F₀ is a force scale, not an additional measured force.

The dimensionless positions are R_A = (−3, 0), R_B = (3, 0), and R_C = (0, 4). Both A–C and B–C separations are 5L₀; A–B is 6L₀. These are lengths in position space. A later head-to-tail force drawing uses force units, not centimetres.

The complete signed data below are the problem inputs. Equal x and y length scales are needed to represent the geometry faithfully. Schematic charge circles do not specify the physical radii of real objects.

Find each force on C

From A to C the displacement is (3, 4)L₀. The charge product is +4q₀², so A repels C along this vector. In force units, F_C←A/F₀ = 4(3, 4)/5³ = (12, 16)/125. Both components are positive.

From B to C the displacement is (−3, 4)L₀. The charge product is −q₀², so B attracts C, reversing that displacement. Therefore F_C←B/F₀ = −(−3, 4)/5³ = (3, −4)/125. Its x component is positive and its y component negative.

The magnitudes are 4F₀/25 and F₀/25, but these two positive numbers cannot simply be added to obtain the resultant magnitude. The directions are different. Write the components before combining them.

Add the arrows, not the magnitudes

Superposition means adding the force vectors from all the other specified charges at the same configuration. Here F_C/F₀ = (15, 12)/125. Both forces push the resultant to the right; their upward and downward components partly oppose each other.

For a head-to-tail construction, draw the A-on-C vector from (0, 0) to (12, 16)/125 in force coordinates. Translate the B-on-C vector without rotating or stretching it, placing its tail at that first head. Its endpoint is (15, 12)/125. The resultant joins the original tail to the final endpoint.

The resultant magnitude is |F_C|/F₀ = √(15² + 12²)/125 = 3√41/125, approximately 0.154. Its unit direction is (5, 4)/√41; its first-quadrant angle satisfies tan θ = 4/5. Direction and magnitude are separate parts of the answer.

If a charge changes position, or a conductor redistributes its charge, calculate the new configuration. Superposition does not say that a contribution remains unchanged when the physical source distribution changes.

Check reciprocity and the complete system

For the static pair law, exchanging recipient and source reverses dᵢⱼ but not qᵢqⱼ. Hence Fᵢ←ⱼ = −Fⱼ←ᵢ. The forces are equal and opposite even if the two masses differ. Their accelerations need not have equal magnitudes, and neither is specified while supports hold the charges.

For all three charges, the net internal force sums to zero. Each pair is central, so its combined torque about any common origin is also zero: (rᵢ − rⱼ) × Fᵢ←ⱼ = 0. The individual net forces and torques need not vanish; their totals provide useful independent checks.

These cancellations are checks on our closed list of electrostatic pair interactions, not a statement that the electric force on each supported object is zero. Beyond electrostatics, do not assume simultaneous matter-only pair forces exhaust electromagnetic momentum accounting. The field can carry momentum too.

Keep materials and geometry in the problem

The point model is useful when the source distribution is specified and the point approximation is justified. A separation large compared with each object’s size often gives a useful leading approximation, but a small net charge can make internal structure or induced contributions comparatively important.

Do not replace every extended object by its total charge at its geometric centre. Particular symmetries can justify such reductions outside a source, but a general distribution requires more information. A neutral object with separated positive and negative regions is an immediate warning against reasoning from total charge alone.

Replacing ε₀ by a material permittivity is not a universal microscopic force rule. For an unbounded homogeneous linear isotropic medium, a constant permittivity can describe the macroscopic electrostatic field of prescribed free sources. That does not automatically give the microscopic or mechanical force on an object embedded in the medium. Interfaces, cavities, finite shapes, anisotropy and nonlinear response require the corresponding material and boundary problem. Those are developed with fields and dielectrics rather than hidden inside a changed constant.

Return path: sum small charge elements

For a specified continuous distribution, replace source charges by dq. A line density λ has units C/m and dq = λ dℓ; a surface density σ has units C/m² and dq = σ dA; a volume density ρ has units C/m³ and dq = ρ dV. Integrate the signed density with its actual length, area or volume element.

For a point recipient q at r separated from the source support, the volume form is F(r) = kₑ q ∫ ρ(r′)(r − r′)/|r − r′|³ dV′. It is the same vector addition applied to infinitesimal contributions. Both vector direction and geometric measure belong inside the integral.

This expression does not supply a point charge’s self-force, and its simple separated-source interpretation does not settle singular, coincident or embedded microscopic source problems. The next field and potential lessons develop spatial descriptions and boundary methods; this return path explains the bridge, not an unrelated collection of solved field geometries.

Return path: follow charge through a boundary

For a fixed volume V, let ρ be charge per volume and j the conventional charge-current density in A/m². With outward unit normal n, charge conservation takes the form d/dt ∫_V ρ dV = −∮_∂V j·n dA. Positive outward current decreases the charge inside.

When the fields are sufficiently regular, the local statement is ∂ρ/∂t + ∇·j = 0. Local continuity says more than keeping one distant total constant: any local change is balanced by charge flow through the boundary. Negative carriers moving inward can correspond to a positive outward conventional current.

Continuity is a conservation constraint, not a complete equation for how rapidly charge moves. Conductivity, geometry, fields, initial conditions and other constitutive information are still needed for a dynamical prediction.

Return path: propagate small input uncertainties

An exact calculation within an ideal model is not an uncertainty-free measurement. For the nonzero force magnitude F = kₑ |q₁q₂|/r², small changes that keep charge signs fixed give dF/F = dkₑ/kₑ + dq₁/q₁ + dq₂/q₂ − 2 dr/r.

For uncorrelated input estimates and a valid first-order approximation, the relative standard uncertainty obeys (u_F/F)² ≈ (u_k/kₑ)² + (u_q₁/q₁)² + (u_q₂/q₂)² + 4(u_r/r)². The minus sign in a sensitivity coefficient does not make its independent variance contribution negative.

For correlated inputs, use u_F² ≈ ΣₐΣᵦ (∂F/∂xₐ)(∂F/∂xᵦ) cov(xₐ, xᵦ), with x = (kₑ, q₁, q₂, r). The double sum already includes both off-diagonal orders. This is a local standard-uncertainty model, not an automatic worst-case bound or a guarantee near a zero charge or a poorly known separation. A vector force also has directional uncertainty.

State what must change beyond electrostatics

Coulomb’s law here describes stationary classical point sources in vacuum. If sources change slowly over a region of size ℓ, an electrostatic approximation needs a relevant change time much longer than ℓ/c, as well as control of motion and material effects. The light-travel comparison is a useful necessary check, not a complete error estimate.

Rapid changes and moving charges require electromagnetic fields, magnetic forces and retardation. At atomic scales, a classical picture of prescribed point trajectories is not enough to predict quantum states or binding. In a material, screening and charge redistribution can change an effective interaction without abolishing the underlying electromagnetic theory.

Electrostatic reasoning remains useful in particle manipulation, sensing and material handling, provided the object and environmental model is stated. Neither a textbook point force nor a static diagram certifies that an actual apparatus is safe. This lesson makes no live-electrical handling recommendation.

Use a question-led solving check

Ask: Which object receives the force? What are the charge signs? What is the source-to-recipient displacement? Are the units consistent? Which assumptions make the point and vacuum models appropriate?

Then calculate each contribution, add components, and recover magnitude and direction. Check dimensions, attraction or repulsion, inverse-square scaling under fixed inputs, and reciprocal pairs. A calculation that passes one check can still fail another.

On a first reading, follow the marked three-charge case and the explained examples. Return to continuous densities, local continuity and uncertainty after the basic vector sum is clear. The later field, potential and material lessons extend this same mechanism rather than requiring a new meaning for charge.

Worked examples

Try it yourself

1. Can charge be cancelled by rotating coordinate axes?

   Show the answer and reasoning

   No.

   Charge is a signed scalar; equal opposite charge amounts can sum to zero without a vector direction.

2. Does zero net charge imply no charged constituents?

   Show the answer and reasoning

   No.

   Positive and negative constituents can sum to zero; their separation can still affect electric interactions.

3. What is 1 nC in coulombs?

   Show the answer and reasoning

   10⁻⁹ C.

   The prefix nano denotes one billionth. Square the conversion when a charge product contains two nanocoulomb factors.

4. An initially neutral object loses electrons. Which sign does it acquire?

   Show the answer and reasoning

   Positive.

   It has lost negative charge. Charge transferred to another object must be included when checking the combined system.

5. Does 1 C by itself specify a current?

   Show the answer and reasoning

   No.

   Current also requires a rate or a time dependence; the unit identity C = A·s does not supply that information.

6. With the same assigned point charges, what happens to the force magnitude when separation doubles?

   Show the answer and reasoning

   It becomes one quarter.

   The ratio is (r₀/(2r₀))² = 1/4, for positive separation and the unchanged point model.

7. What happens if a negative charge product has already reversed the displacement vector?

   Show the answer and reasoning

   Do not apply another attraction minus sign.

   The signed vector expression has already selected the physical direction.

8. For B at (3, 0)L₀ and C at (0, 4)L₀, in which direction is the attractive force on C?

   Show the answer and reasoning

   Right and downward, toward B.

   The source-to-recipient displacement is (−3, 4)L₀; the negative product reverses it.

9. Why is the resultant on C upward although B attracts it downward?

   Show the answer and reasoning

   A’s upward component is larger.

   The y components are 16/125 and −4/125 in force units, leaving +12/125.

10. Is the resultant magnitude 5F₀/25?

    Show the answer and reasoning

    No.

    That adds the two individual positive magnitudes. The vectors are not parallel in the same direction; component addition gives 3√41 F₀/125.

11. Does a more massive charged object exert a larger Coulomb pair force on the lighter object than it receives?

    Show the answer and reasoning

    No.

    The static pair law is reciprocal. Different masses affect acceleration, not this equality of pair-force magnitudes.

12. Does a nonzero calculated electric force prove that a supported charge accelerates?

    Show the answer and reasoning

    No.

    The total force includes the support; the assigned static configuration requires external constraints.

13. Does attraction to small neutral paper prove that the paper originally carried opposite net charge?

    Show the answer and reasoning

    No.

    Polarization in a nonuniform electric field can produce attraction while the object remains neutral.

14. May the point-charge sum include the recipient as its own source?

    Show the answer and reasoning

    No.

    The pair expression is undefined at zero separation and does not give a point charge’s self-force.

15. For a fixed volume, what does a positive net outward conventional current imply?

    Show the answer and reasoning

    The enclosed charge decreases.

    The outward-flux convention enters with a minus sign in dQ_V/dt = −∮ j·n dA; follow signed charge rather than just counting particles.

16. Can independent quadrature be used unchanged for strongly correlated input estimates?

    Show the answer and reasoning

    No.

    The sensitivity-weighted covariance terms are then part of the first-order combined variance. Small-uncertainty and model-validity conditions still apply.

What to carry forward

• Charge is a signed scalar; force is a vector. Net neutrality and absence of charged constituents are different statements.
• For separated static point charges in vacuum, a consistent source-to-recipient displacement and signed charge product determine the force without a second sign rule.
• At fixed source inputs, inverse-square scaling concerns force magnitude. Multiple-source forces must still be added as vectors.
• The supplied three-charge result is (15, 12)F₀/125. The equal-and-opposite pair checks do not make every supported object force-free.
• Conservation, material response, measurement uncertainty and the limits of electrostatics are part of choosing the right model—not optional excuses after a calculation.