SCIENCE · INVESTIGATE AND EXPLAIN

Predict attraction and repulsion between charges

Use charge signs and inverse-square distance scaling to interpret a force pair.

Learning outcome: Apply Coulomb’s law with units and distinguish third-law partners from net force on one object.

Starting recap

Positive and negative are charge signs, not labels for greater or smaller mass. Equal-and-opposite interaction forces act on different objects.

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Try

If two charges exert opposite forces, why does each still need a support to remain fixed?

Open the experiment

Predict

At Q = +2 nC and r = 0.20 m, predict force direction for q = +2 nC and q = −2 nC. Predict the magnitude change if separation doubles.

Your prediction and reason

Experiment

Capture q = +2 and −2 nC at 0.20 m. Keep q fixed and compare 0.40 m. Inspect both force rows and explain the shared signed scale.

Separate electric force, field and potential

Worked teaching model · synthetic values · Two prescribed stationary point charges in vacuum, Q = +2 nC at x = 0 and q = −4 to +4 nC to its right. Rounded Coulomb constant k = 8.99 × 10⁹ N m²/C². The field and potential shown at q are due to Q alone and exclude self-fields. Q and q remain fixed as sources for each selected scenario; controls select comparisons, not time evolution or a charge-transfer process. Mutual electric forces are equal and opposite but act on different charges. External supports that hold positions are not drawn; their forces would balance the electric forces. Potential and pair interaction energy are zero at infinite separation. Work compares the same q moved from r = 0.20 m to the selected r with Q fixed; external work assumes quasistatic transfer with no kinetic-energy change. No acceleration, radiation, polarization, finite-size charges, breakdown or dielectric response. Charge zero is a neutral ideal point with no polarization, not a general model of neutral matter.

Charge separation: 0.2 m Second charge: 2 nC

Reset keeps your prediction, captures and practice history.

Force, field, potential and interaction energy
Quantity	Value	Interpretation
Electric field from Q at q	449.500000 N/C	Rightward; defined even when q = 0
Force on q	0.899000 µN	qE; sign gives direction
Force on Q	-0.899000 µN	Equal-and-opposite partner on a different charge
Potential from Q at q	89.900000 V	Energy per charge; 1 V = 1 J/C
Pair interaction energy	0.179800 µJ	qV, with zero at infinite separation

Work from reference separation 0.20 m
Quantity	Value	Meaning
Reference potential	89.900000 V	From Q at 0.20 m
Initial pair energy	0.179800 µJ	Same q as selected now
Change in pair energy	0.000000 µJ	Final minus initial
Work by electric field	0.000000 µJ	Negative of interaction-energy change
Quasistatic external work	0.000000 µJ	Equal to energy change when kinetic energy does not change

The work comparison holds the selected q fixed between the reference and final positions. Changing the charge slider selects another scenario; it is not a simulated charging process. Supporting forces are omitted from the force-pair diagram, so its electric force values are not the total forces on the held charges.

repulsion: force on q 0.899000 µN; field from Q 449.500000 N/C to the right.

Potential from Q 89.900000 V; pair interaction energy 0.179800 µJ. From 0.20 m, field work 0.000000 µJ and quasistatic external work 0.000000 µJ. Changing q does not change the displayed field or potential from the prescribed Q.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Predict the direction first

Two positive charges repel. Put Q = +2 nC at the origin and q = +2 nC 0.20 m to its right: the electric force on q points right. Reverse q to −2 nC and that force points left, toward Q. The source charge feels the opposite force in both cases. The two forces act on different objects, so they do not cancel in either object’s force balance.

Before adjusting the controls, draw both arrows. Capture each sign of q at the same distance, then double the separation while keeping charge fixed. Explain direction and magnitude separately.

Charge is conserved when it moves

A nanocoulomb (nC) is 10⁻⁹ coulomb (C). Positive and negative describe charge sign, not whether an object has a positive or negative mass. If three electrons move from initially neutral object A to initially neutral object B, A has charge +3e and B has −3e, where e ≈ 1.602 × 10⁻¹⁹ C is the magnitude of one electron’s charge. Their combined charge remains zero.

The slider selects different prescribed net charges; it does not simulate electrons moving between these objects. Its coarse nanocoulomb values are classical approximations, not a particle count.

Calculate a signed force

For the geometry here, with q to the right of Q, F on q = kQq/r². F is the rightward component in newtons (N), r is separation in meters, and k = 8.99 × 10⁹ N m²/C² is the rounded vacuum Coulomb constant. A negative result means a leftward force. The magnitude is k|Qq|/r². In other geometries, direction must be determined from the line joining the charges; this signed scalar rule assumes this particular axis arrangement.

At the default settings, F = (8.99 × 10⁹)(2 × 10⁻⁹)(2 × 10⁻⁹)/(0.20)² = 8.99 × 10⁻⁷ N = 0.899 µN. A micronewton is 10⁻⁶ N. At 0.40 m, the square of distance is four times larger, so force is 0.22475 µN, one quarter as large. Reversing q reverses these signs without changing magnitudes.

The force plots share a fixed scale. Check that doubling distance really shrinks both arrows rather than merely changing the plot’s zoom. Explain why doubling charge and doubling distance together would halve the magnitude.

Held charges are not accelerating charges

The diagram holds both charges in place. External supports, not drawn, would balance their electric forces. The displayed electric pair therefore is not the total force on each held object. Mass, acceleration and a future trajectory cannot be inferred from these static settings.

At q = 0 the ideal point has zero electric force. Real neutral objects can polarize and interact with charged objects; this model excludes polarization. It also excludes finite-size charge distributions, dielectric materials and electrical breakdown. The singular point-charge formula at r = 0 is outside the controls.

Transfer to different charges

Outside the slider’s fixed-source experiment, suppose Q = +3 nC, q = −2 nC and r = 0.30 m. Predict direction, calculate force, and identify the force partner before opening the answer.

Check the charge-force transfer

F on q = (8.99 × 10⁹)(3 × 10⁻⁹)(−2 × 10⁻⁹)/(0.30)² = −5.99333 × 10⁻⁷ N, approximately −0.599333 µN. The force is leftward and attractive. The force on Q from q is +0.599333 µN and acts on Q, not on q.

OpenStax: Coulomb’s law develops the point-charge force and its vector direction.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

State which object each force acts on, how charge and distance affect magnitude, and why changing the charge slider is not a conservation-of-charge simulation.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Explain the broad central band from a single slit Next: Separate electric field from force on a charge

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