
Physics II course sequence · Interactive lesson
Electric field & the test charge
Coulomb's law hands you a force only once two charges are in place. The field asks a different question: what is this point ready to do to any charge brought there? Here you learn to answer that with a probe, and to keep the probe from spoiling the answer it gives.
UPII-02.01 · Lesson 1 of 8 in this unit. Position is not a completion score.Build the model
Divide the force by the probe charge, and what is left belongs to the point.
The key idea
The electric field at a point is the force a charge would feel there divided by that charge, taken in the limit of a vanishingly small positive test charge, so that the result describes the point and not the probe.
Example. A probe of +2.0 nC at P is pulled east with 5.0 × 10⁻⁵ N, so E = 5.0 × 10⁻⁵ N ÷ 2.0 × 10⁻⁹ C = 2.5 × 10⁴ N C⁻¹ east. Swap in +6.0 nC and the force triples to 1.5 × 10⁻⁴ N, but the ratio does not move.
The complete physical model and assumptions
Coulomb's law describes a pair: two charges, one force. The field model splits that statement in two. Fix the charges already in place and call them the sources; then at every point around them ask what force a small positive charge q₀ would feel if it were put there, and divide that force by q₀. Experiment licenses the division, because with the sources held still the force on a probe is strictly proportional to its charge: double q₀ and F doubles, leaving the ratio untouched. So F/q₀ is not a property of the probe at all. It is a vector belonging to the point, defined whether or not anything is there to feel it, and measured in newtons per coulomb. Once that vector is mapped the pair problem is over: the force on any charge later placed there is F = qE, one multiplication, with the sign of q deciding whether the force runs along E or against it. The model has one honest weakness, and it is written into the definition. A probe pushes back on the sources with an equal and opposite force, and if they are free to move — charge on a conductor always is — they rearrange, so what you measure is the field of a distribution your own measurement created. Hence the limit, E = lim(q₀→0) F/q₀: make the probe small enough that the disturbance falls below your tolerance. Charge is quantized, so the limit can never actually be reached; it is an idealisation the experiment approaches, not a procedure it performs.
1From a pair of charges to a property of a point
Coulomb's law is a statement about a pair: put one charge here and another there, and a force appears that is proportional to both of them. That is awkward bookkeeping once a dozen charges sit on a plate, because every new charge restarts the whole calculation. The field model breaks the statement in half. First, freeze the charges already present and call them the sources. Second, walk through the space around them with a small positive probe and record, at every point, the force per unit charge. That record is a vector at each point — a field — and it stands whether or not a probe happens to be there. The payoff is locality: to find the force on a charge placed at P you no longer need to know where the sources sit or how far away they are. You need one vector, the field at P, and the charge itself.
2Why dividing by the probe charge is legitimate
Dividing by q₀ is allowed because measurement says the force is strictly proportional to the probe's charge, so long as the sources stay put. Place +1.0 nC at P and read 3.0 × 10⁻⁵ N; replace it with +3.0 nC at the same point and the reading becomes 9.0 × 10⁻⁵ N; use −2.0 nC and the magnitude is 6.0 × 10⁻⁵ N, reversed. In every case F/q₀ = 3.0 × 10⁴ N C⁻¹. Coulomb's law already contains that proportionality, since the probe's charge enters to the first power, and superposition carries it to any number of sources: each source contributes a force proportional to q₀, and a sum of terms each proportional to q₀ is itself proportional to q₀. The ratio therefore drops the probe out of the answer, which is precisely what a definition of a property of the point requires. Had F gone as q₀², the ratio would still depend on the probe and no field could be defined this way.
3A positive probe sets the direction, by convention
The definition names a positive test charge, and that choice — convention, not physics — fixes the direction. E points the way a positive probe would be pushed. Put a negative probe at the same point and it is dragged the opposite way, yet nothing about the field has changed; only the sign of q in F = qE has. The two vectors are therefore parallel or antiparallel and never at any other angle, because q is a scalar and multiplying E by it can only stretch the vector and possibly reverse it. Concretely: if E at P is 2.0 × 10⁴ N C⁻¹ pointing east, a +5.0 nC charge at P feels 1.0 × 10⁻⁴ N east and a −5.0 nC charge feels 1.0 × 10⁻⁴ N west. Both readings report the same eastward field.
4Sources make the field, probes sample it
Two charges in a problem can have two quite different jobs. Sources make the field; the probe samples it. The distinction bites hardest in F = qE, where E must be the field produced by every charge except q itself: a charge exerts no net force on itself, and its own field diverges at its own location, so including it would make the expression meaningless. The field you draw around a source is the field that acts on other charges, never on the source that made it. The roles are not permanent either. The probe is itself a source, and by Newton's third law it pulls or pushes the source charges with a force of equal magnitude. Which charge you call source and which you call probe is a modelling decision about what you want to know — and the next section is about what that decision costs.
5The probe must not disturb what it is measuring
Every measurement of this kind is paid for. The probe attracts or repels the very charges that produce the field, with an equal and opposite force, and charges free to move will move; charge on a conductor always can. Take a source sphere on a mount that yields 1.0 mm per 100 μN of sideways force, 60.0 mm from a field point where the true field is 1.00 × 10⁵ N C⁻¹. A 2.0 nC probe pushes on it with 2.0 × 10⁻⁹ C × 1.00 × 10⁵ N C⁻¹ = 0.20 mN, which shifts it 2.0 mm. The probe now stands 62.0 mm away, where the field is weaker by the factor (60.0/62.0)² = 0.937, so the reading comes in about 6% low — an error the measurement created. Cut the probe to 0.20 nC and the shift is 0.20 mm and the error 0.7%. The disturbance shrinks with q₀, which is why the definition is written as a limit and not as a plain division.
6Units, and a sense of how big a field can be
The unit follows from the definition: newtons per coulomb. One newton per coulomb is also one volt per metre, since a volt is a joule per coulomb and a joule is a newton metre; the second name earns its keep when potential arrives, and the two are interchangeable everywhere. Three values are worth carrying. The fair-weather atmospheric field near the ground is about 1.0 × 10² N C⁻¹ pointing downward. Dry air breaks down into a spark near 3 × 10⁶ N C⁻¹, which caps the field any bench apparatus can hold. And at one Bohr radius from a proton, 5.29 × 10⁻¹¹ m, the field is 5.1 × 10¹¹ N C⁻¹. Nearly ten powers of ten separate the field you are standing in from the one that binds an atom, and knowing where a number falls in that range is the quickest check on an answer.
Key equations
Read the units, assumptions and limits before substituting.
Operational definition
E = lim(q₀→0) F/q₀
Turns one force reading into a property of the point, ready for every charge that arrives later.
F in N, q₀ in C, E in N C⁻¹; F and E are vectors, q₀ a signed scalarForce on a charge in a known field
F = qE
Runs the definition backwards: with the map in hand the force is one multiplication, no source distances needed.
q signed, in C; E in N C⁻¹; F in N, along E when q > 0 and opposite E when q < 0Probe independence
F₁/q₁ = F₂/q₂ = E
The bench test that the ratio belongs to the point: if it fails, the probe is disturbing the sources.
two probes q₁ and q₂ in C at the same point, feeling forces F₁ and F₂ in NThe field unit, twice named
1 N C⁻¹ = 1 V m⁻¹
N C⁻¹ reads as force per coulomb; V m⁻¹ reads the same number as energy per coulomb per metre.
1 V = 1 J C⁻¹ and 1 J = 1 N m, so the two names are one unitReading E from a hanging probe
E = mg tan θ / |q|
Measures a horizontal field with a balance and a protractor; the thread tension cancels and never enters.
m in kg, g = 9.81 m s⁻², θ measured from the vertical, q in C, E in N C⁻¹The smallest probe there is
|q₀| ≥ e = 1.602 × 10⁻¹⁹ C
Why q₀ → 0 is an idealisation: in practice shrink the probe until the disturbance is under your uncertainty.
e is the elementary charge in C; every free charge is an integer multiple of itInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Set the mount to its softest, k = 50 mN m⁻¹, and push the probe to 2.50 nC: the source retreats 5.0 mm and the reported field falls 15% short. Step the probe back to 0.25 nC and the two bars close to within 2%.
Live results
- SEPARATION
- 6.20 cm
- FORCE ON PROBE
- 187 μN
- PROBE READS F/q₀
- 93.5 kN C⁻¹
- READING ERROR
- -6.3 %
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Probe charge q₀: 2.00 nC; Stiffness of the source mount: 100 mN m⁻¹
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The commonest error is to let the probe leak into the answer. A student reads 6.0 × 10⁻⁵ N on a +3.0 nC probe and concludes that a +6.0 nC probe would sit in a stronger field. It would not: the force doubles to 1.2 × 10⁻⁴ N and the field is 2.0 × 10⁴ N C⁻¹ either way, because the field was already there before either probe arrived. The same slip appears as dividing by the source charge instead of the probe charge, and as the belief that a negative probe reverses the field rather than the force. A second error runs the other way: reading lim(q₀→0) F/q₀ as saying the field fades away as the probe shrinks. Both F and q₀ go to zero together and their ratio holds fixed; the limit is there to switch off the probe's disturbance of the sources, not to switch off the field.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A probe carrying +4.0 nC is held at point P and feels a force of 7.2 × 10⁻⁵ N due east. (a) Find the electric field at P. (b) Find the force on a −2.5 nC charge placed at the same point. (c) State what the field at P becomes if the probe is swapped for one carrying +8.0 nC.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- The field is the force per unit charge on the probe: E = F/q₀ = 7.2 × 10⁻⁵ N ÷ 4.0 × 10⁻⁹ C = 1.8 × 10⁴ N C⁻¹.
- The probe is positive, so the force it feels points along E. The field at P is 1.8 × 10⁴ N C⁻¹ due east.
- For the second charge, run the definition backwards: F = |q|E = 2.5 × 10⁻⁹ C × 1.8 × 10⁴ N C⁻¹ = 4.5 × 10⁻⁵ N. The charge is negative, so the force is opposite E, due west.
- Swapping the probe changes nothing at P. The +8.0 nC probe would feel 1.44 × 10⁻⁴ N east, twice as much force, and F/q₀ is still 1.8 × 10⁴ N C⁻¹.
Answer. E = 1.8 × 10⁴ N C⁻¹ east; the force on the −2.5 nC charge is 4.5 × 10⁻⁵ N west; the field is still 1.8 × 10⁴ N C⁻¹ east.
MediumExample 2
A polystyrene bead of mass 0.42 g carrying −3.0 nC hangs from a light insulating thread. A horizontal electric field is switched on and the thread settles 12° from the vertical, the bead displaced towards the east. Find the magnitude and direction of the field, and the tension in the thread.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Three forces act on the bead: the weight mg downwards, the tension T along the thread, and the horizontal electric force F. In equilibrium the components balance separately, T sin θ = F and T cos θ = mg.
- Divide the first by the second to eliminate the tension: F = mg tan θ = 4.20 × 10⁻⁴ kg × 9.81 m s⁻² × tan 12° = 4.120 × 10⁻³ N × 0.2126 = 8.76 × 10⁻⁴ N, directed east.
- Now divide by the size of the charge: E = F/|q| = 8.76 × 10⁻⁴ N ÷ 3.0 × 10⁻⁹ C = 2.9 × 10⁵ N C⁻¹.
- The bead is negative, so the force is opposite the field. The force points east, so the field points west.
- Tension, from the vertical equation: T = mg/cos θ = 4.120 × 10⁻³ N ÷ 0.9781 = 4.2 × 10⁻³ N. It carries no information about the field, which is why eliminating it was the first move.
Answer. E = 2.9 × 10⁵ N C⁻¹ pointing west; T = 4.2 × 10⁻³ N.
HardExample 3
A source sphere carrying +40 nC hangs from a 25 cm insulating thread and has mass 0.80 g. At the point P, 60.0 mm away on the horizontal through the sphere, Coulomb's law gives an undisturbed field of 9.99 × 10⁴ N C⁻¹ pointing away from the sphere. A probe of +1.0 nC is held at P. Estimate the field the probe reports, then repeat for a probe of +0.10 nC.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- The probe feels F = q₀E₀ = 1.0 × 10⁻⁹ C × 9.99 × 10⁴ N C⁻¹ = 9.99 × 10⁻⁵ N, and by Newton's third law it pushes the source away with the same 9.99 × 10⁻⁵ N.
- The hanging sphere is a pendulum: for a small horizontal displacement x the thread supplies a restoring force mgx/L, so the sphere settles at x ≈ FL/(mg) = 9.99 × 10⁻⁵ N × 0.25 m ÷ (8.0 × 10⁻⁴ kg × 9.81 m s⁻²) = 3.18 × 10⁻³ m, that is 3.18 mm.
- Reading F at the undisturbed 60.0 mm overstates that shift a little, because the push eases as the sphere retreats; here the correction is about 0.3 mm, small enough to leave aside in an estimate.
- The separation is now 60.0 + 3.18 = 63.18 mm, and the field there is smaller by the square of the distance ratio: E = 9.99 × 10⁴ N C⁻¹ × (60.0/63.18)² = 9.99 × 10⁴ × 0.9018 = 9.0 × 10⁴ N C⁻¹.
- So the probe reports about 9.0 × 10⁴ N C⁻¹ where the field was 9.99 × 10⁴ N C⁻¹, near 10% low, and every bit of that shortfall was made by the measurement.
- Shrink the probe tenfold and the force and the shift fall with it: x = 0.318 mm, the separation is 60.32 mm, and E = 9.99 × 10⁴ N C⁻¹ × (60.0/60.32)² = 9.88 × 10⁴ N C⁻¹, now 1.1% low. Cutting q₀ by ten cut the error by about nine, which is the limit in the definition earning its place.
Answer. The +1.0 nC probe reports about 9.0 × 10⁴ N C⁻¹, near 10% below the true 9.99 × 10⁴ N C⁻¹; the +0.10 nC probe reports 9.88 × 10⁴ N C⁻¹, only 1.1% low.
Quick check
Select an answer and explain it before checking.
Review & continue
The electric field at a point is the force a charge would feel there divided by that charge, taken in the limit of a vanishingly small positive test charge, so that the result describes the point and not the probe.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.