Electric Fields: Study Guide (Physics C: E&M 8.3)

Course note. This guide is for the calculus-based Physics C: Electricity and Magnetism course, Topic 8.3. The ideas also appear in the algebra-based Physics 2 course, but here you will add field vectors by components and derive symbolic results that lead into the integrals of Topic 8.4.

Constants used throughout: k = 1/(4πε₀) = 8.99 × 10⁹ N·m²/C², ε₀ = 8.85 × 10⁻¹² C²/(N·m²) and e = 1.60 × 10⁻¹⁹ C. 1 nC = 10⁻⁹ C.

What an electric field is

In Topic 8.1 you used Coulomb’s law to find the force between two charges. A field gives a second way to think about the same force. A charged object (the source) changes the space around it. Any other charge placed in that space then feels a force. The electric field describes that change at each point.

To measure the field at a point, place a small test charge q₀ there and measure the force on it. The field is the force per unit charge:

E = F/q₀

• E is a vector. It has the direction of the force on a positive test charge.
• Its unit is the newton per coulomb, N/C.
• A test charge must be small enough that its presence does not noticeably change the field. A large test charge would push or pull the source charges (for example, the free electrons in a metal source) into new positions, so you would measure a different field.

Once you know E, you can find the force on any charge q placed at that point:

F = qE

If q is positive, F points along E. If q is negative, F points opposite to E. The field itself does not change when you swap the test charge.

Where does the field point? Away from an isolated positive charge, and towards an isolated negative charge. This follows from the force on a positive test charge: it is repelled by a positive source and attracted by a negative one.

The field of a point charge

Put a test charge q₀ a distance r from a point charge q. Coulomb’s law gives F = kqq₀/r². Divide by q₀:

E = kq/r² = q/(4πε₀r²), directed radially away from q if q > 0 and towards q if q < 0.

In vector form, E = (kq/r²) r̂, where r̂ is the unit vector pointing from the source to the point. If q is negative, the minus sign flips the direction.

This is a functional dependence you can use to predict changes without a full calculation:

• E ∝ q. Doubling the source charge doubles the field.
• E ∝ 1/r². Doubling the distance makes E four times smaller; tripling it makes E nine times smaller.
• E does not depend on q₀.

Adding fields: superposition

When several charges are present, each one produces its own field as if the others were not there. The net field at a point is the vector sum of the individual fields:

E_net = E₁ + E₂ + E₃ + …

A reliable method:

1. Draw the point and each source charge. For each charge, draw an arrow at the point: away from a positive charge, towards a negative one.
2. Find each magnitude with kq/r², using the distance from that charge to the point.
3. Split each arrow into x- and y-components, using the geometry of the diagram.
4. Add the x-components and add the y-components separately.
5. Combine: |E| = √(E_x² + E_y²), with direction θ = tan⁻¹(E_y/E_x), checking the quadrant.

Look for symmetry first. If two equal charges sit symmetrically about a point, some components cancel exactly. You can then skip those components entirely. The course limits force calculations (Topic 8.1) to four or fewer charged objects, or more only in highly symmetric cases, so expect field sums of a similar size.

Showing a field: maps and field lines

Because E is a vector at every point in space, you need a picture that shows many vectors at once. There are two common ones.

A vector field map draws an arrow at each point of a grid. The arrow points along E at that point, and its length shows the magnitude. Figure 1(a) is a map for one positive charge: the arrows point outward and shrink quickly with distance, because E ∝ 1/r².

A field-line diagram is a simplified version of the map. You join the arrows into continuous lines. The rules:

• At any point, E is tangent to the field line through that point, in the direction of the arrowhead.
• Lines start on positive charges and end on negative charges (or run off to a great distance).
• The number of lines leaving or entering a charge is proportional to the size of its charge.
• Where lines are close together, the field is strong; where they spread out, it is weak.
• Field lines never cross. At a crossing point the field would have two directions, which is impossible.

In Figure 1(b), +2q has sixteen lines and −q has eight. So only half of the lines from +2q can end on −q; the other half leave the picture. Far away, the pair looks like a single charge +q.

Fields of charged conductors and insulators

What happens to the field depends on whether charge can move.

Conductors in electrostatic equilibrium. In a conductor, free charges move whenever there is a field inside to push them. “Equilibrium” means charges are no longer moving. So:

• E = 0 everywhere inside the conducting material. If it were not zero, free charges would still be moving.
• Any excess charge sits on the surface. Like charges push one another as far apart as they can, which means onto the outer surface.
• Just outside the surface, E is perpendicular to the surface. If E had a component along the surface, it would push surface charges sideways, so they would not be in equilibrium.

Insulators. In an insulator, charges cannot move freely. Excess charge can stay wherever it was put: throughout the inside as well as on the surface. So the field inside a charged insulator can be non-zero.

Spheres. Outside any isolated sphere whose charge is spread with spherical symmetry (a charged metal sphere, or an insulating sphere with charge spread evenly or in shells), the field is the same as that of a point charge with the same net charge placed at the centre: E = kQ/r² for r ≥ R. Inside, the two cases differ: zero inside the metal, but generally non-zero inside the insulator. You will prove these results with Gauss’s law in Topic 8.6, and study conductors in more detail in Unit 10.

Measuring a field

You can design an experiment straight from the definition E = F/q₀. One possible plan for the field near a charged sphere:

1. Mount the charged sphere on an insulating stand, away from other objects and from your hands.
2. Charge a small probe (a tiny conducting ball on an insulating rod) and measure its charge q₀.
3. Attach the probe to a sensitive force sensor. Measure the force F at several distances r from the sphere’s centre, along one line.
4. Calculate E = F/q₀ at each point.
5. Test-charge check. Halve q₀ and repeat one reading. If F halves and F/q₀ stays the same, the probe is small enough not to disturb the sphere’s charge.
6. To test E ∝ 1/r², plot E against 1/r². A straight line through the origin supports the model, and its slope is kQ.

Worked example 1: from force to field and back

Question. A test charge q₀ = +2.0 nC at point P feels a force of 3.6 × 10⁻⁴ N directed north. (a) Find E at P. (b) Find the force on a −5.0 nC charge placed at P instead. (c) The field is produced by one point charge 0.30 m due south of P. Find its charge. (d) Find E at 0.60 m from the source.

1. (a) E = F/q₀ = (3.6 × 10⁻⁴ N) ÷ (2.0 × 10⁻⁹ C) = 1.8 × 10⁵ N/C, north. The test charge is positive, so E has the same direction as F.
2. (b) |F| = |q|E = (5.0 × 10⁻⁹ C)(1.8 × 10⁵ N/C) = 9.0 × 10⁻⁴ N, south. The charge is negative, so the force is opposite to E. The field at P has not changed.
3. (c) E points north, away from a source that is south of P, so the source is positive. From E = kQ/r²: Q = Er²/k = (1.8 × 10⁵)(0.30)² ÷ (8.99 × 10⁹) = +1.8 × 10⁻⁶ C.
4. (d) Doubling r divides E by 2² = 4: E = 4.5 × 10⁴ N/C.

Check. The source is about 900 times larger than the test charge, so q₀ = 2.0 nC is reasonably treated as a test charge. Units: N ÷ C gives N/C.

Worked example 2: net field from two charges

Question. A charge q₁ = +3.0 nC is at the origin and q₂ = −4.0 nC is at (0.40 m, 0). Find the net field at P = (0, 0.30 m), and the force on a proton placed at P.

1. Distances. r₁ = 0.30 m. r₂ = √(0.40² + 0.30²) = 0.50 m.
2. Magnitudes. E₁ = (8.99 × 10⁹)(3.0 × 10⁻⁹) ÷ (0.30)² = 299.7 N/C. E₂ = (8.99 × 10⁹)(4.0 × 10⁻⁹) ÷ (0.50)² = 143.8 N/C.
3. Directions. E₁ points straight up (+y), away from q₁. E₂ points from P towards q₂, along the unit vector (0.40, −0.30)/0.50 = (0.80, −0.60).
4. Components. E₁ = (0, 299.7) N/C. E₂ = (0.80 × 143.8, −0.60 × 143.8) = (115.1, −86.3) N/C.
5. Add. E_net = (115.1, 213.4) N/C. |E_net| = √(115.1² + 213.4²) = 242 N/C, at tan⁻¹(213.4/115.1) = 61.7° above the +x direction.
6. Force on a proton. F = eE = (1.60 × 10⁻¹⁹ C)(242 N/C) = 3.9 × 10⁻¹⁷ N, in the same direction as E_net.

Check. Adding the magnitudes would give about 440 N/C, far too big: the y-components partly cancel. An electron at P would feel the same size of force in the opposite direction.

Worked example 3: a field from a graph

Question. A student uses a field meter to measure E at distances r from the centre of a small charged sphere. Use the data to test whether E ∝ 1/r² and to find the sphere’s charge.

r (m)	0.10	0.15	0.20	0.25	0.30
E (N/C)	4520	1980	1130	710	505
1/r² (m⁻²)	100	44.4	25.0	16.0	11.1

1. Choose axes. The model is E = kQ/r². Writing x = 1/r² gives E = (kQ)x, a straight line through the origin with slope kQ. So plot E (vertical) against 1/r² (horizontal).
2. Plot and fit. The points lie close to a straight line through the origin (Figure 3). That supports E ∝ 1/r².
3. Slope. Using the line, not a single data point: at 1/r² = 100 m⁻² the line is at about 4510 N/C, so the slope is 4510 ÷ 100 ≈ 45 N·m²/C.
4. Charge. kQ = 45 N·m²/C, so Q = 45 ÷ (8.99 × 10⁹) = 5.0 × 10⁻⁹ C (5.0 nC).

Check. From r = 0.10 m to r = 0.20 m, E falls from 4520 to 1130 N/C, a factor of exactly 4, as the inverse-square model predicts. A plot of E against r would be a curve, which makes the slope hard to use.

Common misconceptions

• “The field depends on the test charge.” The force does; the field does not. E is set by the source charges.
• “Field lines are paths that charges follow.” A field line gives the direction of the force, which is the direction of the acceleration, not of the velocity. A charge released on a curved field line does not, in general, follow that line.
• “No line drawn means no field.” Lines are only a sample. The field exists between them.
• Adding magnitudes instead of vectors. Two fields of 300 N/C and 140 N/C can give anything from 160 N/C to 440 N/C, depending on direction. Use components.
• Using the wrong distance. Each kq/r² uses the distance from that charge to the point, not the distance between the charges.
• Wrong direction for a negative source. E points towards a negative charge, whatever the sign of the test charge.
• “The field is zero inside every charged sphere.” That is true for a conductor in equilibrium, not for an insulator with charge spread through it.
• “Charge spreads evenly through a metal.” Excess charge on a conductor sits on its surface.

Where this leads

Next, in Topic 8.4, you will replace sums over a few point charges with integrals over continuous charge distributions, such as rods and rings: see the 8.4 study guide. Gauss’s law (Topic 8.6) will then prove the conductor and sphere results above. Practise now with the practice questions, then use the revision notes and the topic checklist.