PhysSandbox

Electric Field: Point Charges, Field Lines and Superposition

A 45-minute guided lesson: students measure how the field of a point charge falls off with distance, see that fields add as vectors (and can cancel where the potential does not), read field-line diagrams of a dipole, and build a nearly uniform field between two charged plates.

Level: Grades 10–12 · AP Physics 2 · IB Physics HL45 minSimulator used in this lesson: Electric Field Visualizer

Learning objectives

  • State that the field of a point charge falls as 1/r^2 and the potential as 1/r, and verify both with measurements.
  • Apply the superposition principle: the field of several charges is the vector sum of their individual fields.
  • Read a field-line diagram: direction of the field, relative strength from line density, and where the field is zero.
  • Distinguish electric field (a vector) from electric potential (a scalar) at a point.

Students should already know

  • Coulomb’s law for the force between two point charges.
  • Adding vectors by components.

Standards

  • NGSS HS-PS2-4
  • NGSS HS-PS3-5
  • AP Physics 2 · Unit 1 (Electric Charges, Fields and Potential)
  • IB Physics · Topic D.2
Warm-up

1.A field you cannot see

~5 min

A single positive charge sits alone in space. Imagine placing a tiny positive test charge at different points around it. Before using the simulator, predict:

In the simulator, arrows show the direction and relative strength of the field vec E, gold lines are field lines, dashed curves are equipotentials and the yellow dot is a draggable probe that reports |E| and V at its position. Units are arbitrary, so look at ratios, not absolute numbers.

Explore 1

2.How fast does the field fall off?

~10 min

One positive charge (q = 2) sits at the centre. The probe is placed 0.20 units to the right of it.

Your task

  1. Record |E| at probe and V at probe for the starting distance r = 0.20.
  2. Drag the probe straight to the right so that it is twice as far away (r ≈ 0.40 — the grid helps; the readouts will be about one quarter and one half of the starting values). Record |E| and V again.
  3. Drag it to half the starting distance (r ≈ 0.10). What happened to |E|? To V?
  4. Fill in: when r doubles, |E| is multiplied by ___ and V is multiplied by ___.

At r = 0.20 you should read |E| ≈ 50 and V ≈ 10; at r = 0.40, |E| ≈ 12.5 and V ≈ 5; at r = 0.10, |E| ≈ 200 and V ≈ 20. Doubling the distance divides the field by four and the potential by two: the field follows an inverse-square law, the potential an inverse law.

Field of a point charge

( E = dfrack qr^2, directed away from a positive charge, towards a negative one\)

Potential of a point charge

( V = dfrack qr (a scalar; zero at infinity)\)

Force on a test charge

( vec F = q_0 vec E\)

Explore 2

3.Superposition: fields add, and can cancel

~12 min

Now two equal positive charges (q = 2 each) sit at x = −0.20 and x = +0.20. The probe starts exactly at the midpoint.

Your task

  1. Read |E| at probe and V at probe at the midpoint. Which one is zero? Explain using arrows: what does each charge’s field look like at that point?
  2. Drag the probe slightly up from the midpoint (to y ≈ 0.2). The field is no longer zero — in which direction does it point? Why does the x-component still cancel?
  3. Look at the field lines between the charges. Why do no lines pass through the midpoint, and why do the lines seem to “repel” each other there?
  4. Drag one charge so that it sits closer to the probe. Does the point of zero field move towards the bigger or the smaller gap?

At the midpoint the two fields are equal in size and opposite in direction, so vec E = 0 — but the potentials are both positive and add: V ≈ 20. This is the key difference between the two quantities: E is a vector (contributions can cancel), V is a scalar (contributions from like charges only pile up). Above the midpoint the horizontal components still cancel by symmetry, but both vertical components point up, so vec E points straight up (|E| ≈ 35 at y = 0.2).

Superposition (field)

( vec E = vec E_1 + vec E_2 + dots = sum_i dfrack q_ir_i^2\,hat r_i\)

Superposition (potential)

( V = V_1 + V_2 + dots = sum_i dfrack q_ir_i\)

Explore 3

4.The dipole

~8 min

Replace the right-hand charge by a negative one of the same size: +2.5 at x = −0.18 and −2.5 at x = +0.18. This pair is an electric dipole — the field pattern of a water molecule or a polarised antenna.

Your task

  1. At the midpoint read |E| and V. Compare with the two-positive-charge case: which quantity is now zero and which is large? Explain with arrows.
  2. Follow a field line with your finger: where does it start and where does it end? Do any lines cross?
  3. Drag the probe along the vertical line x = 0 (the perpendicular bisector). What is V everywhere on that line? What does that make the line?
  4. Look at the dashed equipotentials: where are they closest together, and how does that relate to the arrow lengths?

At the dipole’s midpoint both fields point the same way (from + towards −), so |E| ≈ 154 is large, while the potentials +kq/r and −kq/r cancel: V = 0. The whole perpendicular bisector is the V = 0 equipotential. Field lines start on the positive charge, end on the negative one, never cross, and are densest — equipotentials closest together — where the field is strongest, near the charges and in the gap between them.

Explore 4

5.Building a uniform field

~5 min

Seven positive charges in a vertical row on the left and seven negative ones on the right imitate a pair of charged parallel plates — the geometry of a capacitor.

Your task

  1. Move the probe around the middle region between the plates. How much does |E| change? In which direction do the arrows point?
  2. Move the probe to the edge of the plates and then outside them. Where does the “uniform field” idea break down?
  3. Equipotentials between the plates: are they evenly spaced? What does that tell you about how V changes with x?

Between the plates the arrows are nearly parallel and equal in length and the equipotentials are evenly spaced vertical lines: E is approximately uniform and V changes linearly with position, E = Δ V / d. Near the edges the lines bulge outward (the *fringing field*) and the approximation fails.

Worked example

6.Field on the axis of two charges

~5 min

Where is the field zero?

A charge of +4 μC is at x = 0 and a charge of +1 μC is at x = 0.30 m. Find the point on the line between them where the electric field is zero. (k = 8.99 × 10⁹ N·m²/C².)

  1. Between two positive charges the fields point in opposite directions, so they can cancel. Let the point be at distance x from the 4 μC charge, hence 0.30 - x from the 1 μC charge.
  2. Set the magnitudes equal: dfrack · 4x^2 = dfrack · 1(0.30 - x)^2. The constant k cancels — units of charge too, as long as both are the same.
  3. Take square roots: dfrac2x = dfrac10.30 - x Rightarrow 2(0.30 - x) = x Rightarrow x = 0.20\ textm.
  4. Check: the zero is closer to the smaller charge (0.10 m from it, 0.20 m from the larger one), as expected — the weaker charge needs a shorter distance to match the stronger one.

x = 0.20 m from the 4 μC charge (0.10 m from the 1 μC charge). In the simulator, place q = 4 at x = −0.15 and q = 1 at x = +0.15 and drag the probe until |E| is minimal — it should be at x ≈ 0.05.

Quiz

7.Check your understanding

Answer each question, then press Check. Numeric answers from the simulator are accepted within the stated tolerance.

1.You triple your distance from a point charge. The electric field strength becomes…

2.At 0.20 m from a point charge the field is 50 N/C. What is the field at 0.50 m?

N/C±3 % accepted

3.At the midpoint between two equal positive charges…

4.At the midpoint of a dipole (+q and −q)…

5.Which statements about electric field lines are correct? (Select all that apply.)

6.Charges of +9 μC (at x = 0) and +1 μC (at x = 0.40 m) lie on a line. At what x (in m) between them is the field zero?

m±3 % accepted

8.Homework

  1. Place q = +4 at x = −0.15 and q = +1 at x = +0.15. Find the position of zero field by dragging the probe, then check it against the formula from the worked example. How far is it from each charge? Open simulator with these settings ↗
  2. A quadrupole: +2 at (−0.2, 0), +2 at (0.2, 0), −2 at (0, −0.2), −2 at (0, 0.2). Sketch the field-line pattern first, then compare with the simulator. Where is E zero? What is V there? Open simulator with these settings ↗
  3. In the parallel-plate preset, select the middle charge of one plate and delete it. Describe what happens to the uniform field. Then explain why real capacitors use continuous metal plates rather than rows of point charges. Open simulator with these settings ↗

Notes for teachers

  • Pacing: Warm-up 5 min, Explore 1 10 min, Explore 2 12 min, Explore 3 8 min, Explore 4 5 min, worked example and quiz 5 min. Explore 4 can be dropped if the class has not met capacitors yet.
  • Field and potential are in arbitrary units (k = 1, charge and distance in simulator units). Every task is built on ratios, which are unit-independent. If students ask, 0.2 simulator units can be read as 0.2 m with charges in units of 1/k.
  • Probe positions set by hand will not be exact; the readouts in the text (50 / 12.5 / 200) are for exactly 0.20 / 0.40 / 0.10. Values within ~10 % are fine for the ratio argument — the grid spacing helps students place the probe.
  • Misconceptions targeted: (1) E and V confused — Explore 2 and 3 are built as a matched pair (E = 0, V ≠ 0 vs E ≠ 0, V = 0); (2) “field lines are paths of charges” — field lines give the force direction, not the trajectory, which is worth saying aloud at Explore 3; (3) “more lines means more charge *there*” — line density indicates field strength at that point.
  • Extension: ask students to find the zero-field point for two charges of opposite sign and unequal magnitude (it lies outside the pair, on the side of the smaller charge).
  • Numeric quiz answers accept ±3 %.