Scivra

Preview

Electric Field Lines

Visualize fields from point charges on a 2D field map

Coulomb's law told you the force between two charges. But here is the question that bothered physicists for decades: how does one charge even KNOW the other one is there? Nothing touches, nothing stretches between them — yet the force arrives. Faraday's answer, sketched in the 1830s before Maxwell gave it mathematics, is the idea this lab makes visible: a charge doesn't reach across space, it ALTERS the space itself. Every point around a charge carries a vector — the electric field — and any other charge placed there simply reads the local instructions: force = charge × field. That reframe is more than bookkeeping. It lets you price the field once, from the sources alone, and then predict the force on any test charge you like. And it obeys one beautifully simple accounting rule, superposition: each source writes its own vectors into space as if the others didn't exist, and the total at any point is just the vector sum — direction included, which is where your intuition will trip. Two equal positive charges, and exactly halfway between them the field is not twice as strong but precisely ZERO, because the two contributions point in opposite directions and annihilate. Flip one charge negative and that same midpoint doubles instead. In this lab the field lines draw themselves live as you dial the charges: they stream out of the red positive sphere, dive into the blue negative one, thin out with the inverse square, and bend around the null point where the vectors cancel. The panel prices the field and the potential at the origin on every frame. Your first job is to predict that cancellation before you touch anything — then go watch space itself keep the books.

What you'll be able to do

  • Run the superposition sign test at the midpoint with the panel's own numbers: at defaults (+5/−5 ×10⁻⁹ C, separation 2 m — charges at x=±2, the HTML's placement) each charge contributes k·5×10⁻⁹/2² = 11.2375 N/C, and for the dipole both contributions point the SAME way (+x, toward the negative charge) so |E| at origin reads 22.48 N/C; flip charge2 to +5 and the same two 11.2375 N/C vectors point in OPPOSITE directions and the row collapses to exactly 0.00 N/C — 'the vector sum of all these fields' (OpenStax §5.4 verbatim), direction included
  • Certify the inverse square on the midpoint field: holding the default dipole, sweep separation 0.5 → 2 → 4 m and watch |E| at origin fall 359.60 → 22.48 → 5.62 N/C — an ×8 distance divides the field by exactly 8² = 64 (E = 2k|q|/sep² at the midpoint), the 1/r² from Coulomb's law carried unchanged into the field concept
  • Read the field-line picture by its own rules (OpenStax §5.6): lines leave the red positive sphere and land on the blue negative one, never cross ('the field was pointing in two different directions at a single point… obviously impossible'), and pack tightest near each charge where the field is strongest — then check the signed ledger underneath: the V row at the midpoint of the symmetric dipole reads exactly 0.0 V because kq₁/r + kq₂/r keeps signs, where an unsigned sum would read 44.95 V

Formulas

E⃗=keqr2r^\vec{E} = k_e \frac{q}{r^2}\hat{r}
Electric field from a point charge
ke=8.99×109 N⋅m2/C2k_e = 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2
Coulomb's constant
E⃗total=∑iE⃗i\vec{E}_{total} = \sum_i \vec{E}_i
Superposition principle
F=ke∣q1q2∣r2F = k_e \frac{|q_1 q_2|}{r^2}
Coulomb's law (force between charges)
V=keqrV = k_e \frac{q}{r}
Electric potential from a point charge
E⃗=−∇V\vec{E} = -\nabla V
Field is negative gradient of potential

Make a prediction

Default state: charge1 = +5 and charge2 = −5 (×10⁻⁹ C) sit at x = ±2 m, and the |E| at origin row reads 22.48 N/C — each charge contributes 11.2375 N/C, both pointing toward the negative side. Now flip charge2 to +5 (same magnitude, same positions). What does the |E| at origin row read?

No grading here — commit to a guess, then scroll down and test it yourself.

This is a Pro experiment

Upgrade to Pro to access this experiment — or keep learning with one of our free labs.

View Pro plansBrowse free experiments
Scroll for the debrief ↓

Check

Did the lab agree with you?

Your prediction

You skipped the prediction — jump back to the preview and commit to a guess first; the comparison is the whole point.

Answer: 0.00 N/C — the two contributions are equal in magnitude but now point in OPPOSITE directions at the midpoint, and cancel exactly

Superposition is a VECTOR sum — 'the total electric field, then, is the vector sum of all these fields' (OpenStax §5.4, verbatim). Each +5 ×10⁻⁹ C charge contributes 11.2375 N/C at the midpoint; with both positive, the field from the charge at x=−2 points +x (away from it) while the field from the charge at x=+2 points −x (away from it too) — equal magnitudes, opposite directions, exact annihilation: 0.00 N/C. OpenStax's own Example 5.4 computes this geometry and watches the components 'cancel each other out'; the sim draws it as the field lines bending away from a dead point between the spheres. Option B forgets that direction is half the data — same magnitudes, opposite vectors, zero sum. Option C is the documented scalar-superposition error — adding strengths and discarding directions (Campos et al. 2019 traces field-line misreadings to exactly this move); note the trap's symmetry: for the DIPOLE the two contributions really did reinforce (22.48), because there the directions happened to agree. Direction decides whether equal contributions double or vanish — that is the whole lesson.

Quiz (0/5)

At the midpoint between a +5 nC and −5 nC charge separated by 2 m, what is the direction of the electric field?

Two equal positive charges: where is the electric field zero?

Calculate the force between +5 nC and −5 nC charges separated by 2 m.

Two +3 nC charges and one −6 nC charge form a triangle. Describe the field at the center.

Why are equipotential surfaces always perpendicular to field lines? Give a physical argument.

You can now

  • Run the superposition sign test at the midpoint with the panel's own numbers: at defaults (+5/−5 ×10⁻⁹ C, separation 2 m — charges at x=±2, the HTML's placement) each charge contributes k·5×10⁻⁹/2² = 11.2375 N/C, and for the dipole both contributions point the SAME way (+x, toward the negative charge) so |E| at origin reads 22.48 N/C; flip charge2 to +5 and the same two 11.2375 N/C vectors point in OPPOSITE directions and the row collapses to exactly 0.00 N/C — 'the vector sum of all these fields' (OpenStax §5.4 verbatim), direction included
  • Certify the inverse square on the midpoint field: holding the default dipole, sweep separation 0.5 → 2 → 4 m and watch |E| at origin fall 359.60 → 22.48 → 5.62 N/C — an ×8 distance divides the field by exactly 8² = 64 (E = 2k|q|/sep² at the midpoint), the 1/r² from Coulomb's law carried unchanged into the field concept
  • Read the field-line picture by its own rules (OpenStax §5.6): lines leave the red positive sphere and land on the blue negative one, never cross ('the field was pointing in two different directions at a single point… obviously impossible'), and pack tightest near each charge where the field is strongest — then check the signed ledger underneath: the V row at the midpoint of the symmetric dipole reads exactly 0.0 V because kq₁/r + kq₂/r keeps signs, where an unsigned sum would read 44.95 V

What you'll learn

  • Field Line Rules. Electric field lines start on positive charges and end on negative charges. They never cross — if they did, a test charge at the crossing point would have two directions to go, which is physically impossible. The density of lines represents field strength.
  • Coulomb's Law. The electric force between two point charges follows an inverse-square law, just like gravity. Double the distance and the force drops to one-quarter. Double either charge and the force doubles.
  • Superposition Principle. When multiple charges are present, the total electric field at any point is the vector sum of the individual fields. This means you calculate each charge's contribution independently, then add them as vectors — direction matters!
  • Equipotential Surfaces. An equipotential surface connects all points at the same electric potential. No work is done moving a charge along an equipotential. These surfaces are always perpendicular to field lines — this is a powerful geometric tool for visualizing fields.
  • Dipole Fields. A pair of equal and opposite charges (a dipole) creates a characteristic field pattern: lines arc from the positive charge to the negative charge. Far from the dipole, the field falls off as 1/r³ — faster than a single charge. Water molecules are permanent dipoles, which is why water is such a good solvent.

Step-by-step

  1. Adjust the two charges and their distance from center.
  2. The field lines all lie in one flat plane — the scene itself is three-dimensional, so use your mouse to orbit and view the map from any angle.
  3. Try +/+ (repulsion), +/− (attraction), and equal charges, then load the Three Charges preset to break the symmetry with an off-axis third charge.

Key formulas

  • E⃗=keqr2r^\vec{E} = k_e \frac{q}{r^2}\hat{r}Electric field from a point charge
  • ke=8.99×109 N⋅m2/C2k_e = 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2Coulomb's constant
  • E⃗total=∑iE⃗i\vec{E}_{total} = \sum_i \vec{E}_iSuperposition principle
  • F=ke∣q1q2∣r2F = k_e \frac{|q_1 q_2|}{r^2}Coulomb's law (force between charges)
  • V=keqrV = k_e \frac{q}{r}Electric potential from a point charge
  • E⃗=−∇V\vec{E} = -\nabla VField is negative gradient of potential

Frequently asked questions

At the midpoint between a +5 nC and −5 nC charge separated by 2 m, what is the direction of the electric field?
The correct answer is: Toward the negative charge. Each charge contributes a field at the midpoint. Add them as vectors — they point in the same direction here.
Two equal positive charges: where is the electric field zero?
The correct answer is: At the midpoint between them. By symmetry, the fields cancel at the midpoint. Set q₁=q₂=+5 and observe.
Calculate the force between +5 nC and −5 nC charges separated by 2 m.
The correct answer is: 5.6 × 10⁻⁸ N. F = kq₁q₂/r². Use k = 8.99×10⁹.
Two +3 nC charges and one −6 nC charge form a triangle. Describe the field at the center.
The correct answer is: Points toward the −6 nC charge. You can work it out this way: use superposition. Enable the third charge and set q₁=q₂=+3, q₃=−6 in a triangular configuration.
Why are equipotential surfaces always perpendicular to field lines? Give a physical argument.
The correct answer is: Because no work is done along an equipotential, requiring E⊥displacement. Moving along an equipotential does no work (ΔV=0). Work = qE·d = 0 requires E⊥d.