
Physics II course sequence · Interactive lesson
Electric field lines
Lines are how a field gets onto paper. Learn the four habits that keep the drawing honest — tangent, density, endpoints, no crossings — and you can read a charge ratio, a null point and a conductor boundary straight off a diagram.
UPII-02.04 · Lesson 4 of 8 in this unit. Position is not a completion score.Build the model
Every rule for drawing the lines comes from a property of E.
The key idea
An electric field line is a curve drawn so that the electric field is tangent to it at every point, arrowed in the direction of the force on a positive test charge.
Example. An isolated +2.0 nC charge drawn at 8 lines per nanocoulomb gets 16 radial lines. At r = 0.10 m they pierce a sphere of area 0.126 m² at 127 lines m⁻², where E = 1.8 × 10³ N C⁻¹.
The complete physical model and assumptions
A field line is a curve drawn so that E is tangent to it at every point, and the whole picture follows from that definition plus one counting convention: pick n lines per coulomb, give every charge N = n|q| of them, aimed outward from positive charge and inward to negative. The rest are consequences, not extra rules. Lines cannot cross, because superposition gives one field vector at each point and so one tangent direction. Their crowding measures the strength, because the n|q| lines leaving a charge all pierce every sphere around it, and the 1/r² thinning of the count is exactly the 1/r² of Coulomb's law — with any other exponent, lines would have to stop in empty space. Lines meet a conductor at right angles, because a component of E along the surface would push the surface charge sideways until it was gone. And the model stops where the drawing does: n is a free choice, so no absolute field strength can be read off a picture; a printed diagram is a plane slice of a three-dimensional pattern, so density on the page falls as 1/r rather than 1/r²; and no line is defined through a point where E = 0. A line shows where the force points, never where a released charge will go.
1A line is a curve that follows the field
The field E already exists at every point; a line is a way of walking through it. Start anywhere, read E, take a short step along it, read E again at the new point, and repeat. The curve that comes out satisfies dy/dx = Ey/Eₓ everywhere along itself, and its arrowhead points the way E points, which is the direction of the force on a positive test charge and the opposite of the force on a negative one. Two things follow at once. Through a given point there is exactly one such curve, because there is exactly one field vector there — which is what makes the whole picture well defined. And the curve is not a thing: start from a neighbouring point and you get a neighbouring line, and the field in the gap between two drawn lines is every bit as real as the field on them.
2Crowding measures strength because the field falls as 1/r²
Fix a scale: n lines per coulomb, so a charge q carries N = n|q| of them. Around an isolated charge every one of those lines pierces every sphere drawn about it, so the count per unit area at radius r is n|q|/(4πr²) while the field is |q|/(4πε₀r²). Both carry the same 1/r², so eliminating q leaves N/A⊥ = n ε₀ E: the crowding is the field strength, in a fixed ratio. Take q = 3.0 nC drawn at 6 lines per nanocoulomb, so N = 18. At r = 0.15 m the sphere has area 0.2827 m², giving 63.7 lines m⁻², and E = 1.20 × 10³ N C⁻¹. Move out to r = 0.30 m: the area is four times larger, the density falls to 15.9 lines m⁻², and E falls to 300 N C⁻¹. The match is not luck. Only an inverse-square law lets unbroken lines carry the field outward without any of them having to stop in empty space.
3Lines start on positive charge and end on negative charge
Because the count belongs to the charge, one n must serve every charge in a diagram, and then the counts are in proportion to the charges. At 4 lines per nanocoulomb, 20 lines leave a +5.0 nC charge and 12 arrive at a −3.0 nC charge nearby. Twelve of the twenty find the negative charge; the remaining eight have nowhere to land and run out of the picture. That leftover is set by the net charge alone, N_∞ = n|q₁ + q₂| = 4 × 2.0 = 8, which is why a pair seen from far away looks like a single charge of +2.0 nC. The same bookkeeping works for any closed surface you care to draw: lines out minus lines in equals n qenc, since nothing begins or ends in empty space. That statement becomes Gauss's law once the counting is replaced by an integral.
4Two lines cannot cross, and a null point is the exception
At a crossing, a curve would have two tangents, so E would have two directions at the same point. Superposition forbids it: the contributions from every charge add to one vector, with one magnitude and one direction. Note what is being added — the field vectors, not the pictures. You cannot lay one charge's line pattern over another's and call the overlay a field map. The single exception is a point where the vector sum is zero, because a zero vector has no direction. For +4q and +q a distance d apart, that point sits on the line between them at 2d/3 from the larger charge, since 4/x² = 1/(d − x)² gives 2(d − x) = x. Lines approaching it swerve aside, and the boundary lines that divide the pattern — one charge's territory from the other's — do meet there, at a definite angle. It is the one place in the picture where the no-crossing rule has nothing to say.
5At a conductor, every line arrives at 90° and stops
In electrostatic equilibrium the field inside conducting material is zero, so no line is drawn in the metal at all. At the surface the parallel component must vanish too: any Eₜ would push the free surface charge sideways, and the charge keeps moving until it has arranged itself so that Eₜ = 0. What survives is perpendicular, so external lines strike a conductor at right angles and terminate on the surface charge they induce. The surface count follows the charge: n|σ| lines leave or land per square metre. Set that against N/A⊥ = n ε₀ E, which applies just outside where the lines run perpendicular, and n cancels to leave E = σ/ε₀ immediately outside any conductor. On a pear-shaped conductor, σ is largest at the sharp end, so that is where the lines crowd and where the outside field is strongest.
6Reading a diagram, and what the flat page distorts
A picture will honestly give you signs from the arrowheads, a charge ratio from the counts, the direction of E at a point from the tangent, and a ranking of strengths from the crowding. It will not give you a number for E, because n was your choice. It also lies about one thing quietly. A printed pattern is a plane slice through a three-dimensional one, and lines drawn radially on paper thin out as 1/r along a circle rather than 1/r², so the page always understates how fast the field really dies away. Use the drawing for structure and ranking, and the algebra for values. In the laboratory the order is reversed: a probe maps the equipotential contours, and the field lines are then drawn crossing them at right angles, pointing towards lower potential. Even measured, a field line is inferred and never photographed.
Key equations
Read the units, assumptions and limits before substituting.
Tangent condition for a line
dy/dx = Ey/Eₓ
Step along E from any starting point and you have traced one line. The tangent is a direction, never a path.
x and y in m along the drawn curve; Eₓ and Ey are the field components in N C⁻¹ (= V m⁻¹)Line density fixes the field strength
N/A⊥ = n ε₀ E
Crowding is not a metaphor: count the lines through a small perpendicular window and you have measured E.
A⊥ in m², held perpendicular to E; n is the diagram's lines per coulomb, in C⁻¹; ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²Lines belong to charge
N = n |q|
One n serves the whole diagram, so two charges' line counts stand in the ratio of their charges.
q in C, n in lines per coulomb; lines run outward from q > 0 and inward to q < 0Net count through a closed surface
Nₒᵤₜ − Nᵢₙ = n qenc
Line counting is Gauss's law before the integral: only enclosed charge changes the net count.
qenc is the net charge enclosed by the surface, in C; the counts are pure numbersLines that escape to infinity
N_∞ = n |q₁ + q₂|
Far from any group of charges the pattern settles into that of a single charge of the net amount.
q₁ and q₂ in C are the two charges of a pair; the remaining lines run from one to the otherConductor boundary condition
Eₜ = 0 at a conductor surface
Lines arrive at 90° and stop on the surface charge, at n|σ| lines per square metre of surface.
in electrostatic equilibrium; Eₜ is the field component parallel to the surface, in N C⁻¹, and σ is the surface charge density in C m⁻²Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Set qB to −1: the arrows between the charges point straight along the axis, from A to B. Push it to +1 and those same arrows swing vertical, because the two contributions now cancel along the axis instead of adding — and the tangent angle at P jumps with them.
Live results
- TANGENT ANGLE AT P
- 29 ° from +x
- LINES ON B
- 8 lines
- A's LINES ENDING ON B
- 8 lines
- LINES TO OR FROM ∞
- 0 lines
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 on B, in units of qA: −1.00; Probe P, distance right of A: 2.0 cm; Probe P, height above the axis: 1.4 cm
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 read a field line as a track and expect a released charge to run along it. The field gives force, so it fixes acceleration, not velocity. A charge released from rest does start off along the line, but the moment it has speed, inertia carries it straight on while the line curves away; only a line that is straight by symmetry is also a path. Two smaller slips travel with it. The blank space between drawn lines is not field-free — the lines are a sample, and doubling n doubles every count without changing any field. And a line is not a stretched string: it has no tension, no length, and nothing moves along it.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A diagram of two charges is drawn throughout at 3 lines per nanocoulomb. Charge A is +8.0 nC and charge B is −2.0 nC. How many lines are drawn on each, how many run from A to B, how many leave the picture, and how many net lines cross a large sphere enclosing both?
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 count belongs to the charge, with one n for the whole diagram: NA = n|qA| = 3 lines nC⁻¹ × 8.0 nC = 24 lines, drawn outward because A is positive.
- On B, NB = 3 lines nC⁻¹ × 2.0 nC = 6 lines, drawn inward because B is negative.
- Nothing starts or ends in empty space, so every line arriving at B began on a charge, and A is the only source: all 6 of B's lines come from A. That leaves 24 − 6 = 18 of A's lines with nowhere to land, and they run off to infinity.
- Check against the net charge: N_∞ = n|qA + qB| = 3 × 6.0 = 18 outward, the same 18. A large sphere encloses qenc = +6.0 nC, so lines out minus lines in = n qenc = 18.
Answer. 24 lines out of A and 6 into B; 6 of them connect the pair, 18 escape to infinity, and any closed surface around both has a net 18 lines leaving it.
MediumExample 2
Charges q₁ = +6.0 nC at the origin and q₂ = +2.0 nC at (0.100 m, 0) lie in a plane. Find the direction of the field line through P = (0.040 m, 0.030 m), and say where that line ends up. Take k = 8.99 × 10⁹ N m² C⁻².
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.
- From q₁ to P: r₁ = √(0.040² + 0.030²) = 0.050 m along the unit vector (0.80, 0.60). E₁ = kq₁/r₁² = 8.99 × 10⁹ × 6.0 × 10⁻⁹ ÷ 0.00250 = 2.158 × 10⁴ N C⁻¹, so its components are (1.7261 × 10⁴, 1.2946 × 10⁴) N C⁻¹.
- From q₂ to P: the displacement is (−0.060, 0.030) m, so r₂ = 0.0671 m and the unit vector is (−0.894, 0.447). E₂ = 8.99 × 10⁹ × 2.0 × 10⁻⁹ ÷ 0.00450 = 3.996 × 10³ N C⁻¹, with components (−3.574 × 10³, 1.787 × 10³) N C⁻¹.
- Superpose the vectors, never the two line patterns: Eₓ = 1.7261 × 10⁴ − 0.3574 × 10⁴ = 1.369 × 10⁴ N C⁻¹ and Ey = 1.2946 × 10⁴ + 0.1787 × 10⁴ = 1.473 × 10⁴ N C⁻¹.
- The line's tangent is that sum: dy/dx = Ey/Eₓ = 1.473/1.369 = 1.08, so the line runs at 47.1° above the +x axis — steeper than the 36.9° of q₁ alone, because q₂ pushes P upward and to the left.
- Both charges are positive, so the diagram has nowhere for a line to end: this one escapes to infinity, and far out it joins a pattern indistinguishable from a single +8.0 nC charge.
Answer. The field line through P has tangent 47° above the +x axis, slope dy/dx = 1.08, where |E| = 2.0 × 10⁴ N C⁻¹; the line escapes to infinity.
HardExample 3
A point charge q = +4.0 nC sits at the centre of a neutral conducting spherical shell of inner radius 2.0 cm and outer radius 3.0 cm. The diagram is drawn at 5 lines per nanocoulomb. Count the lines in each region, find the two induced surface charge densities, and recover the field just outside the shell from the line density alone.
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.
- N = n|q| = 5 lines nC⁻¹ × 4.0 nC = 20 lines leave the central charge and cross the cavity radially.
- The metal is in electrostatic equilibrium, so E = 0 inside it and no line may be drawn there: all 20 stop on the inner surface. Lines end only on negative charge, so that surface holds −20/5 = −4.0 nC over 4π(0.020 m)² = 5.027 × 10⁻³ m², giving σᵢₙ = −7.96 × 10⁻⁷ C m⁻² and 3.98 × 10³ lines m⁻² arriving.
- The shell is neutral, so +4.0 nC is left on the outer surface; 20 lines start there, perpendicular to it, over 4π(0.030 m)² = 1.131 × 10⁻² m². So σₒᵤₜ = +3.54 × 10⁻⁷ C m⁻² and the count is 20 ÷ 0.01131 = 1.77 × 10³ lines m⁻².
- Read the field off that count with N/A⊥ = n ε₀ E, taking n = 5 × 10⁹ lines C⁻¹: E = 1.77 × 10³ ÷ (5 × 10⁹ × 8.85 × 10⁻¹²) = 4.0 × 10⁴ N C⁻¹ just outside the shell.
- Two checks: σₒᵤₜ/ε₀ = 3.54 × 10⁻⁷ ÷ 8.85 × 10⁻¹² = 4.0 × 10⁴ N C⁻¹, and kq/b² = 8.99 × 10⁹ × 4.0 × 10⁻⁹ ÷ 0.030² = 4.0 × 10⁴ N C⁻¹. Outside, the 20 lines look exactly as they would with the shell removed.
Answer. 20 lines cross the cavity, none the metal, 20 leave the outside; σᵢₙ = −0.80 μC m⁻² and σₒᵤₜ = +0.35 μC m⁻², and E just outside is 4.0 × 10⁴ N C⁻¹.
Quick check
Select an answer and explain it before checking.
Review & continue
An electric field line is a curve drawn so that the electric field is tangent to it at every point, arrowed in the direction of the force on a positive test charge.
- 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.