
Physics II course sequence · Interactive lesson
Fields of Point Charges
Once the field is defined, the first source worth evaluating is a single point charge. Every later result in this unit is built by adding copies of this one, so learn to read it as a picture: a direction fixed by r̂, a size fixed by 1/r², and a sign that reverses the arrows without touching their lengths.
UPII-02.02 · Lesson 2 of 8 in this unit. Position is not a completion score.Build the model
One charge, one rule: radial direction, inverse-square size
The key idea
The electric field of a point charge is a vector defined at every point in space, directed along the line from the charge to that point, with magnitude k|q|/r² — outward if the charge is positive, inward if it is negative.
Example. A 1.0 nC charge sets up 9.99 × 10³ N C⁻¹ at 3.0 cm from it. Move out to 6.0 cm and the field is not halved but quartered, to 2.50 × 10³ N C⁻¹.
The complete physical model and assumptions
Put a charge q somewhere and divide Coulomb's law by the test charge you used to feel it: what is left mentions the test charge nowhere. E⃗ = (1/4πε₀)(q/r²) r̂, with r measured from q and r̂ the unit vector pointing from q towards the place you are asking about. Two claims are packed into that line. The direction is radial everywhere, because an isolated point charge has no axis of its own to prefer, and the sign of q — not the sign of whatever you bring to probe it — decides whether the arrows point out or in. The magnitude is the inverse square, k|q|/r², so every factor of ten further out costs a factor of a hundred in field, and on logarithmic axes the whole law is a straight line of slope −2. The model then has two edges. Push r towards zero and E diverges, which is the model objecting that a genuinely sizeless charge is a fiction; real charge sits on objects with extent, and outside a spherical one the point formula is exact right down to the surface. Go the other way and the model improves: any bounded lump with net charge q settles onto kq/r² once you are a few times its own size away, which is why the humblest formula in electrostatics also describes a charged sphere, a charged balloon and a star.
1Divide Coulomb's law by the test charge
Coulomb's law gives the force on a test charge q₀ sitting a distance r from q: F⃗ = k q q₀ r̂/r². Divide by q₀ and the probe vanishes from the result, leaving E⃗ = k q r̂/r². That division is the whole move, and it is why the field is a property of the source and of the point rather than of whatever happens to be sitting there. Formally the limit q₀ → 0 is taken, so the probe cannot push the source around while it measures. Put numbers through it: for q = 1.0 nC at r = 3.0 cm, E = (8.99 × 10⁹ × 1.0 × 10⁻⁹) ÷ (0.030)² = 8.99 N m² C⁻¹ ÷ 9.0 × 10⁻⁴ m² = 9.99 × 10³ N C⁻¹. Nothing in that calculation knows whether anything is sitting at 3.0 cm.
2r̂ makes the field radial, and the sign of q chooses in or out
The symbol r̂ is not one fixed direction. It is a different unit vector at every field point, always running from the charge outwards to the point you asked about, so the arrows of a single point charge fan out along radii and do nothing else. Keep the sign inside q and the direction looks after itself: q > 0 gives E⃗ along +r̂, pointing away; q < 0 gives −r̂, pointing back towards the charge. The magnitude is k|q|/r² either way. Take q = +1.0 nC at the origin and the point (3.0 cm, 4.0 cm): r = 5.0 cm, so r̂ = (0.60, 0.80) and E = 3.60 × 10³ N C⁻¹, giving E⃗ = (2.16 × 10³, 2.88 × 10³) N C⁻¹. Change the charge to −1.0 nC and both components reverse while the magnitude is untouched.
3The cube in the vector form is not an inverse cube
When the charge sits at r⃗ₛ rather than at the origin, the separation vector is r⃗ − r⃗ₛ and the unit vector is r̂ = (r⃗ − r⃗ₛ)/|r⃗ − r⃗ₛ|. Substituting that into E⃗ = kq r̂/|r⃗ − r⃗ₛ|² produces the cube: E⃗ = kq (r⃗ − r⃗ₛ)/|r⃗ − r⃗ₛ|³. Count the powers before believing the exponent. The denominator carries three powers of length, but the numerator is a vector whose own length is that same distance, so two powers survive and the magnitude still falls as 1/r². Check it arithmetically with a charge at the origin and a field point at (0.060 m, 0.080 m): the numerator has length 0.100 m, the denominator is 0.100³ = 1.00 × 10⁻³ m³, and the ratio is 1.00 × 10² m⁻², which is exactly 1/(0.100 m)².
4Read the inverse square as a ratio, and as a straight line
You rarely need E twice; you need to know how E changed. The ratio form does that without k ever appearing: E₂/E₁ = (r₁/r₂)². For a 1.0 nC charge, moving from 2.0 cm to 4.0 cm takes the field from 2.25 × 10⁴ N C⁻¹ to 5.62 × 10³ N C⁻¹, a factor of exactly four for a factor of two in distance. Take logarithms and the same law becomes log E = log(k|q|) − 2 log r: on log–log axes it is a straight line of slope −2, and changing q slides that line up or down without tilting it, which is a clean way to see that the charge sets the strength while the geometry sets the fall-off. Differentiate instead and you get dE/dr = −2E/r, so dE/E = −2 dr/r: a 1% error in your distance is a 2% error in the field.
5The near edge: where a point stops being a point
As r → 0 the formula sends E to infinity, and that is a statement about the model, not about nature. Real charge lives on objects with size. For a sphere carrying its charge uniformly, the field outside is exactly kQ/r² measured from the centre, so the point model holds right down to the surface and only fails inside — which is where the formula's floor really sits. The surrounding air adds a second floor: dry air breaks down near 3.0 × 10⁶ V m⁻¹, so a 1.0 nC charge cannot be squeezed inside a radius of √(k|q|/E) = √(8.99 × 10⁹ × 1.0 × 10⁻⁹ ÷ 3.0 × 10⁶) = 1.7 mm without sparking. For fundamental particles, classical electrostatics is replaced long before the divergence bites: a proton's field at the Bohr radius, 5.29 × 10⁻¹¹ m, is already 5.15 × 10¹¹ N C⁻¹.
6The far edge, where the model only gets better
Go the other way and the approximation improves. Model a small lump of total charge q as two charges q/2 a distance d apart, and measure on the perpendicular bisector: each contributes k(q/2)/(r² + d²/4), the components across the bisector cancel, and the sum is E = kq r/(r² + d²/4)³⁄² = (kq/r²)(1 + d²/4r²)(−3/2). At r = 10d the true field is 0.37% below the point-charge value, and at r = 6d it is 1.0% below. Measured along the line through the two charges the correction has the opposite sign, 0.75% high at r = 10d, but it shrinks just as fast, as (d/r)². So far means a few times the source's own size, and beyond that a point charge is an excellent model of anything carrying net charge.
Key equations
Read the units, assumptions and limits before substituting.
Field of a point charge
E⃗ = (1/4πε₀)(q/r²) r̂
Magnitude and direction in one line, and no test charge appears — the field belongs to q and to the point you name.
E in N C⁻¹ = V m⁻¹, q in C, r in m; k = 1/4πε₀ = 8.99 × 10⁹ N m² C⁻², with ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻²Charge away from the origin
E⃗ = kq (r⃗ − r⃗ₛ)/|r⃗ − r⃗ₛ|³
The form you actually compute with. Not an inverse cube: one of the three powers only normalises the direction.
r⃗ₛ locates the charge and r⃗ the field point, both in m; E⃗ comes out in N C⁻¹Inverse-square scaling
E₂/E₁ = (r₁/r₂)²
Answers the question you usually have: three times further out is nine times weaker, whatever the charge happens to be.
One charge, two distances in the same unit; the ratio is dimensionlessHow steeply the field falls
dE/dr = −2k|q|/r³ = −2E/r
Says what a positioning error costs: 1% out in r is 2% out in E, and the penalty is worst where r is small.
N C⁻¹ m⁻¹; in fractional form dE/E = −2 dr/rDistance for a required field
r = √(k|q|/E)
Inverts the law — it locates a breakdown radius, or the range at which a detector stops reading.
r in m for q in C and E in N C⁻¹; valid outside the source onlyOutside a spherical charge
E = kQ/r² for r ≥ R
A uniformly charged ball is exactly a point charge outside itself, so the model's near edge is the surface, not r = 0.
Q the total charge in C, R the radius in m, r measured from the centreInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Move r from 1.00 cm to 2.00 cm and watch the field drop to a quarter while the marker stays on the same straight line; then flip the sign slider and see the graph hold still while the inset arrow reverses.
Live results
- FIELD E
- 45.0 kN C⁻¹
- E AT TWICE r
- 11.2 kN C⁻¹
- RADIAL COMPONENT
- 45.0 kN C⁻¹
- SLOPE dE/dr
- -45.0 kN C⁻¹ cm⁻¹
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. Charge magnitude |q|: 2.0 nC; Sign of q (−1 or +1): 1; Probe distance r: 2.00 cm
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The commonest slip here is reading E⃗ = kq (r⃗ − r⃗ₛ)/|r⃗ − r⃗ₛ|³ as an inverse-cube law. It is not. The numerator is a vector whose length is the very distance in the denominator, so one of those three powers is spent turning r⃗ − r⃗ₛ into a unit vector and only two are left to set the size. Test it: at 2.0 cm from a 1.0 nC charge the field is 2.25 × 10⁴ N C⁻¹, and at 4.0 cm it is 5.62 × 10³ N C⁻¹ — down by four, not by eight. The companion slip is measuring r from the wrong place. It runs from the charge itself: not from the surface of a sphere, not from your chosen origin, and not from some other field point. Get that origin wrong and every direction in the answer is wrong too, because r̂ is built from the same difference.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A small plastic bead carries a charge of −4.5 nC. Treating it as a point charge, find the magnitude and direction of the electric field it produces at a point 2.5 cm away.
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.
- Convert before anything else: q = −4.5 nC = −4.5 × 10⁻⁹ C and r = 2.5 cm = 0.025 m. Leaving r in centimetres is where this calculation usually dies.
- The magnitude uses the size of the charge only: E = k|q|/r² = (8.99 × 10⁹ × 4.5 × 10⁻⁹) ÷ (0.025)² = 40.5 N m² C⁻¹ ÷ 6.25 × 10⁻⁴ m² = 6.5 × 10⁴ N C⁻¹.
- The sign sets the direction and nothing else: q is negative, so E⃗ lies along −r̂ — radially inward, towards the bead, at every point on the 2.5 cm sphere around it.
- Scale check against a benchmark: 1.0 nC at 1.0 cm gives 9.0 × 10⁴ N C⁻¹. Here the charge is 4.5 times larger and the distance 2.5 times greater, so 9.0 × 10⁴ × 4.5/6.25 = 6.5 × 10⁴ N C⁻¹. Agreed.
Answer. E = 6.5 × 10⁴ N C⁻¹, directed radially inward, towards the bead.
MediumExample 2
A point charge q = +6.0 nC sits at the origin. Find the electric field at P = (8.0 cm, 6.0 cm): its magnitude, its x and y components, and the angle it makes with the x axis.
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.
- Distance from the charge to P: r = √(0.080² + 0.060²) = √(0.0064 + 0.0036) = √0.0100 = 0.100 m.
- Magnitude: E = kq/r² = (8.99 × 10⁹ × 6.0 × 10⁻⁹) ÷ (0.100)² = 53.9 N m² C⁻¹ ÷ 1.00 × 10⁻² m² = 5.4 × 10³ N C⁻¹.
- Direction: r̂ = r⃗/r = (0.080, 0.060)/0.100 = (0.80, 0.60), which is a unit vector since 0.80² + 0.60² = 1.00.
- Components: Eₓ = 5394 × 0.80 = 4.3 × 10³ N C⁻¹ and Ey = 5394 × 0.60 = 3.2 × 10³ N C⁻¹. Both are positive because q is positive and P lies up and to the right of the charge.
- Angle: θ = atan(0.60/0.80) = 36.9° above the x axis — the same angle as the line from the charge to P, which is what radial has to mean.
Answer. E⃗ = (4.3 × 10³, 3.2 × 10³) N C⁻¹: magnitude 5.4 × 10³ N C⁻¹, at 36.9° above the x axis.
HardExample 3
A metal sphere of radius 1.5 cm carries positive charge spread uniformly over its surface. Dry air breaks down at 3.0 × 10⁶ V m⁻¹. (a) Find the largest charge the sphere can hold in air. (b) With that charge on it, how far from the centre has the field fallen to 1.0 × 10³ N C⁻¹? (c) Over what change in distance does the field there vary by 1%?
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.
- Outside a uniformly charged sphere the field is exactly that of a point charge at the centre, so the strongest field anywhere is at the surface: E(R) = kQ/R².
- (a) Set E(R) = 3.0 × 10⁶ V m⁻¹ and invert: Q = E R²/k = (3.0 × 10⁶ × 0.015²) ÷ (8.99 × 10⁹) = 675 ÷ (8.99 × 10⁹) = 7.5 × 10⁻⁸ C, that is 75 nC.
- (b) Invert the same law for r instead: r = √(kQ/E) = √((8.99 × 10⁹ × 7.508 × 10⁻⁸) ÷ 1.0 × 10³) = √(0.675 m²) = 0.82 m.
- Sanity check the model at that range: 0.82 m is 55 sphere radii, so the point-charge form is good to far better than the two significant figures quoted.
- (c) Use dE/E = −2 dr/r. A 1% change needs |dr| = 0.005r = 0.005 × 0.82 m = 4.1 mm. At the sphere's surface the same 1% takes only 0.005 × 0.015 m = 0.075 mm.
Answer. (a) 7.5 × 10⁻⁸ C, or 75 nC. (b) r = 0.82 m. (c) 4.1 mm there, against just 0.075 mm at the surface — the field is far flatter far out.
Quick check
Select an answer and explain it before checking.
Review & continue
The electric field of a point charge is a vector defined at every point in space, directed along the line from the charge to that point, with magnitude k|q|/r² — outward if the charge is positive, inward if it is negative.
- 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.