Electric Fields: Study Guide (Physics 2 10.3)

What an electric field is

In Topic 10.1 you met Coulomb’s law: two charged objects push or pull on each other even when they do not touch. The field model explains this action at a distance in two steps.

1. A charged object (the source) sets up an electric field in the space around it.
2. Any other charge placed in that space feels a force from the field at its own location.

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

E = F_E / q

A test charge is a point charge small enough that it does not noticeably change the field around it. A large charge would pull or push the source charges and move them. The unit of E is the newton per coulomb (N/C). In Topic 10.5 you will see that the same unit can be written as V/m.

This is the same idea as the gravitational field from Physics 1, g = F_g/m, measured in N/kg. Mass sets up a gravitational field; charge sets up an electric field.

Example. A test charge of +2.0 nC at point P feels a force of 6.0 × 10⁻⁵ N to the east. The field at P is E = (6.0 × 10⁻⁵ N) ÷ (2.0 × 10⁻⁹ C) = 3.0 × 10⁴ N/C, to the east. Now put a −4.0 nC charge at P instead (with the source charges held in place). The force on it is F_E = qE, with magnitude (4.0 × 10⁻⁹ C)(3.0 × 10⁴ N/C) = 1.2 × 10⁻⁴ N, to the west.

Direction of the field and of the force

The electric field is a vector. Its direction is defined by the force on a positive test charge.

• Near an isolated positive charge, the field points away from it (a positive test charge is repelled).
• Near an isolated negative charge, the field points toward it (a positive test charge is attracted).
• A positive charge in a field feels a force in the same direction as E.
• A negative charge (for example an electron) feels a force opposite to E.

So F_E = qE works as a vector equation when you keep the sign of q. A negative q flips the direction.

Field of a point charge

Put a test charge q₀ a distance r from a point charge q. Coulomb’s law gives a force of size k|q||q₀|/r². Divide by |q₀|:

E = k|q| / r², where k = 1/(4πε₀) = 9.0 × 10⁹ N·m²/C²

Notice that q₀ has cancelled. The field depends only on the source charge and the distance from it.

The field follows an inverse-square law. For a +5.0 nC point charge:

Distance r	E = k q / r²
0.30 m	500 N/C
0.60 m	125 N/C
0.90 m	55.6 N/C

Double the distance and E falls to 1/4 of its value. Triple the distance and E falls to 1/9.

Adding fields: superposition

When several charges are nearby, the net field at a point is the vector sum of the fields that each charge would make on its own. The course asks you to calculate fields from up to four charged objects, or more when the arrangement is highly symmetric. A reliable method:

1. Draw the point, and draw an arrow for the field from each charge (away from positives, toward negatives).
2. Find each magnitude with E = k|q|/r². Use the distance from that charge to the point.
3. Split each field into x- and y-components. Use symmetry to spot components that cancel.
4. Add the components, then find the magnitude and direction of the net field.

Never add magnitudes when the arrows point in different directions.

Field maps and field lines

There are two common pictures of a field (Figure 1).

A vector field map draws an arrow at many points. Each arrow shows the direction of E at its tail and its length shows the magnitude. Around a single positive charge, all arrows point outward and they shrink quickly with distance.

A field-line diagram is a simplified model of the vector field map. Its rules:

• The field at any point is tangent to the line through that point, in the direction of the arrowheads.
• Lines start on positive charges and end on negative charges (or go off to infinity).
• The number of lines leaving or entering a charge is proportional to the size of the charge.
• Where lines are closer together, the field is stronger.
• Lines never cross, because the field has only one direction at each point.

In Figure 1(b), 16 lines leave +2q but only 8 end on −q, because −q has half the size of charge. The field is strongest close to each charge and in the region between them, where the lines are most crowded. Field lines show the direction of the force on a positive charge, not the path a charge follows: a moving charge can cut across field lines.

Charged conductors and insulators

In a conductor (such as a metal), electrons move freely. In an insulator they cannot. This difference controls where excess charge ends up.

Conductors in electrostatic equilibrium (no charges moving):

• The field inside the conducting material is zero. If there were a field inside, the free electrons would feel a force and move. They keep moving until the field inside has been cancelled. “Equilibrium” means that has already happened.
• Excess charge sits on the outer surface. Like charges repel and spread out as far apart as they can.
• At the surface the field is perpendicular to the surface. A component along the surface would push charges sideways, and they would move until that component vanished.
• A charged conducting sphere acts, for points outside it, like a point charge of the same total charge at its centre. Inside, the field is zero.

Insulators. Charge cannot move freely through an insulator, so excess charge can stay where it was placed: inside the material as well as on its surface. The field inside a charged insulator need not be zero. The course only asks for qualitative reasoning about fields inside insulators. Outside a sphere whose charge is spread with spherical symmetry, the point-charge result still holds, whether the sphere is a conductor or an insulator.

Background (beyond what you must calculate): the zero field inside a conductor is the idea behind electrostatic shielding. A closed metal box with no charge in its cavity keeps outside static fields away from what is inside.

Worked example 1: two charges on a line

Question. A +4.0 nC point charge sits at x = 0 and a −2.0 nC point charge sits at x = 0.30 m. (a) Find the net electric field at x = 0.10 m. (b) Find the force on an electron placed there. (c) Is there any point on the x-axis where the net field is zero?

1. Distances from the point: 0.10 m from the +4.0 nC charge, 0.20 m from the −2.0 nC charge.
2. Field from +4.0 nC: E₁ = (9.0 × 10⁹)(4.0 × 10⁻⁹) ÷ (0.10)² = 3600 N/C, pointing away from it, so in the +x direction.
3. Field from −2.0 nC: E₂ = (9.0 × 10⁹)(2.0 × 10⁻⁹) ÷ (0.20)² = 450 N/C, pointing toward it, also in the +x direction.
4. (a) Both point the same way, so they add: E = 3600 + 450 = 4050 N/C, in the +x direction.
5. (b) F = |q|E = (1.6 × 10⁻¹⁹ C)(4050 N/C) = 6.48 × 10⁻¹⁶ N. The electron is negative, so the force is in the −x direction.
6. (c) Between the charges both fields point in +x, so they cannot cancel. To the left of x = 0 the point is always closer to the larger charge, so its field always wins. The zero must lie to the right of the −2.0 nC charge. There, at position x, set the sizes equal: 4.0/x² = 2.0/(x − 0.30)². Taking square roots, x − 0.30 = x/√2, so x = 0.30 ÷ (1 − 1/√2) = 1.02 m.

Answer. (a) 4.1 × 10³ N/C in +x (4050 N/C). (b) 6.5 × 10⁻¹⁶ N in −x. (c) Yes, at x ≈ 1.02 m.

Check. At x = 1.024 m both fields have size 34.3 N/C and point in opposite directions. The zero is closer to the smaller charge, which makes sense.

Worked example 2: two charges in a plane

Question. Two +3.0 nC charges are fixed at (0, +4.0 cm) and (0, −4.0 cm). Point P is at (3.0 cm, 0). (a) Find the net field at P. (b) The upper charge is replaced by a −3.0 nC charge. Find the new net field at P.

1. Distance from each charge to P: r = √(0.030² + 0.040²) = 0.050 m (a 3-4-5 triangle).
2. Size of each field: E = (9.0 × 10⁹)(3.0 × 10⁻⁹) ÷ (0.050)² = 1.08 × 10⁴ N/C.
3. Components: the line from each charge to P makes cos θ = 0.030/0.050 = 0.6 with the x-axis and sin θ = 0.8. So each field has an x-part of 0.6 × 10 800 = 6480 N/C and a y-part of 0.8 × 10 800 = 8640 N/C.
4. (a) Both charges positive. The upper charge pushes P down and to the right; the lower charge pushes it up and to the right. The y-parts cancel by symmetry. E = 2 × 6480 = 1.30 × 10⁴ N/C in the +x direction.
5. (b) Now the field from the upper charge points toward it: up and to the left. The field from the lower charge still points up and to the right. The x-parts cancel and the y-parts add: E = 2 × 8640 = 1.73 × 10⁴ N/C in the +y direction.

Interpretation. Changing the sign of one charge turns its field round, so a different pair of components cancels. Symmetry saved half the work in both parts. In (b), a proton at P would be pushed in +y, and an electron in −y.

Worked example 3: a charged metal sphere

Question. A solid metal sphere of radius 0.10 m carries +2.0 nC and is in electrostatic equilibrium. Find the field (a) 0.30 m from its centre, (b) just outside its surface and (c) 0.05 m from its centre.

1. (a) Outside, treat the sphere as a point charge at its centre: E = (9.0 × 10⁹)(2.0 × 10⁻⁹) ÷ (0.30)² = 200 N/C, pointing radially outward.
2. (b) Just outside the surface, r = 0.10 m: E = (9.0 × 10⁹)(2.0 × 10⁻⁹) ÷ (0.10)² = 1800 N/C, perpendicular to the surface, outward.
3. (c) The point is inside the metal, so E = 0.

Check. The point in (a) is 3 times as far from the centre as the point in (b), and 1800 ÷ 200 = 9 = 3². All the excess charge is on the surface.

Common misconceptions

• “The field only exists where a charge is placed.” The source charges set up the field everywhere around them. A test charge just detects it.
• “The field depends on the test charge.” The force depends on q; the ratio F_E/q does not. Double the test charge and the force doubles, but E stays the same.
• “The force is always along E.” Only for a positive charge. An electron is pushed opposite to the field.
• Adding magnitudes. Fields are vectors. In Worked example 2(a) the two fields each have size 1.08 × 10⁴ N/C, yet the net field is 1.30 × 10⁴ N/C, not 2.16 × 10⁴ N/C.
• “Field lines are the paths charges follow.” A line shows the direction of the force at each point, not the trajectory.
• “There is no field between the lines.” A diagram draws only a few lines. The field exists at every point; its direction there follows the nearby lines.
• “The field is zero inside every charged object.” That is true only inside a conductor in equilibrium. Inside a charged insulator the field can be nonzero.
• Measuring r to the surface of a sphere. For a charged sphere, measure r from the centre.

Where this leads

The field tells you the force per unit charge. Topic 10.4 asks how much energy is stored when charges are pushed together against these forces: see Electric Potential Energy. Topic 10.5 then defines the potential, which is energy per unit charge. Test yourself now with the practice questions, then use the revision notes and the topic checklist to consolidate.