9478 / 2027
Electric Fields overview

Lesson 1 of 7

Point-charge force and field

  • I can recall and use Coulomb's law for the force between two point charges in free space or air.Syllabus

    Syllabus 9478, 14(a). Recall and use Coulomb's law for the force between two point charges in free space or air.

  • I can recall and use the inverse-square electric field of a point charge in free space or air.Syllabus

    Syllabus 9478, 14(b). Recall and use the inverse-square electric field of a point charge in free space or air.

Make a guess. You are not marked.

A point charge produces an electric field strength of 900 N/C at 0.10 m from it. What is the field strength at 0.30 m from it?

  1. 600 N/C, because the field falls by 300 N/C for every 0.10 m.
  2. 100 N/C, because the field falls with the square of the distance.
  3. 300 N/C, because the distance has become three times as large.
Show the answer

100 N/C, because the field falls with the square of the distance.

For a point charge E = Q / (4 pi epsilon0 r2). Three times the distance divides the field by 32 = 9, so 900 / 9 = 100 N/C.

A source charge creates an electric field. The force on a chosen test charge depends on both that field and the test charge's sign and magnitude.

Use Coulomb's law with a direction

For two stationary point charges Q and q in free space or air, separated by distance r, the force acts along the line joining them. Its magnitude is:

F = |Qq|/(4πε0r2) = K|Qq|/r2
K = 1/(4πε0)

Like signs repel; unlike signs attract. The force on Q and the force on q are equal and opposite, but act on different charges. They do not cancel in a force diagram for either one charge.

Here ε0 is the permittivity of free space. Use the supplied rounded calculation value K = 9.00 × 109 N m2/C2. Charge is in coulombs, C, and r is in metres. One nanocoulomb, nC, is 10-9 C.

Separate field from the test-charge force

Electric field strength is force per unit positive test charge, with direction given by the force on that positive charge. For a point source, its magnitude is K|Q|/r2, directed away from positive Q and towards negative Q.

If outward from the source is chosen positive, the signed radial component and the test-charge force are:

Er = Q/(4πε0r2) = KQ/r2
Fr = qEr

In vector form, F = qE. A positive test charge experiences force along E; a negative one experiences force opposite E. A negative component has a meaning only after the positive direction has been stated.

Estimate the scale before substituting

A few nanocoulombs at a distance of a few tenths of a metre, with the supplied K, gives field of order 102-103 N/C. Acting on a nanocoulomb-scale charge, this suggests force of order 10-6 N. The force is small even though the field's numerical value is hundreds, because their units and quantities differ.

Worked source and test charge

Positive source, negative test charge

A fixed source Q = +4.00 nC has a test charge q = -2.00 nC at r = 0.300 m. Take outward as positive.

Er = (9.00 × 109)(4.00 × 10-9)/(0.300)2
= +400 N/C
Fr = (-2.00 × 10-9)(400)
= -8.00 × 10-7 N

The field is outward, while the force on q is inward, towards Q. Replacing q with +2.00 nC reverses that force to outward, with unchanged magnitude.

Changing the test charge changes its force

Keep the source Q = +4.00 nC and the 0.300 m separation unchanged. Teal E describes the source field at q; purple arrows show the two electric interaction forces. Field and force use different units and separate arrow scales.

Negative test charge: attraction

Negative test charge: attraction with an unchanged outward source fieldSource Q is positive four nanocoulombs at the left and the chosen test charge q is at the right, 0.300 metres from its centre. The test charge is negative two nanocoulombs. Its purple electric force points left, towards Q; the electric force on Q points right. The interaction forces have equal arrow lengths and equal magnitudes, eight times ten to the minus seven newtons, but act on different charges. The separately labelled teal field arrow below q points right in both cases and represents four hundred newtons per coulomb. A guide identifies the point where this field is evaluated; it is not another force or a trajectory. Charge symbols are schematic point markers. Only the electric interaction forces are shown; holding the source fixed may require another force.0.300 mForce on QForce on q+-Q = +4.00 nCq = -2.00 nCEach force: 8.00 × 10-7 NE at q400 N/C outwardThe source field is unchanged.

Positive test charge: repulsion

Positive test charge: repulsion with an unchanged outward source fieldSource Q is positive four nanocoulombs at the left and the chosen test charge q is at the right, 0.300 metres from its centre. The test charge is positive two nanocoulombs. Its purple electric force points right, away from Q; the electric force on Q points left. The interaction forces have equal arrow lengths and equal magnitudes, eight times ten to the minus seven newtons, but act on different charges. The separately labelled teal field arrow below q points right in both cases and represents four hundred newtons per coulomb. A guide identifies the point where this field is evaluated; it is not another force or a trajectory. Charge symbols are schematic point markers. Only the electric interaction forces are shown; holding the source fixed may require another force.0.300 mForce on QForce on q++Q = +4.00 nCq = +2.00 nCEach force: 8.00 × 10-7 NE at q400 N/C outwardThe source field is unchanged.

The equal and opposite forces belong to different charges. They do not cancel in the force diagram for q. The test-charge approximation treats Q's field as unchanged when q is replaced.

The panels compare alternative test charges at the same separation from a fixed positive source. Field and force arrows name different quantities and have different units. Interaction forces act on different charges.

Changing q alone changes its force, while the field due to the fixed source stays the same in this model. The test-charge approximation assumes the test charge does not rearrange the source. Doubling r quarters both the source field and the force on an unchanged q.

Try it step by step

Work it out in steps: Product of the charge sizes in C2, then Force F = Q1 Q2 / (4 pi epsilon0 r2) in N.

  1. Problem 1. Charges of +2.0 x 10-6 C and +3.0 x 10-6 C are 0.30 m apart in air. Take 1 / (4 pi epsilon0) = 8.99 x 109 N m2 C-2. Find the size of the force between them.
  2. Problem 2. Charges of +4.0 x 10-6 C and +5.0 x 10-6 C are 0.20 m apart. Use the same constant. Find the size of the force between them.
  3. Problem 3. Charges of +1.5 x 10-6 C and -6.0 x 10-6 C are 0.15 m apart. Use the same constant. Find the size of the force between them.
Show the answer

Problem 1.

Product of the charge sizes in C2: Q1 Q2 = 2.0 x 10-6 x 3.0 x 10-6 = 6.0 x 10-12 C2

Force F = Q1 Q2 / (4 pi epsilon0 r2) in N: F = 8.99 x 109 x 6.0 x 10-12 / 0.302 = 0.599 N, repulsive

Problem 2.

Product of the charge sizes in C2: Q1 Q2 = 4.0 x 10-6 x 5.0 x 10-6 = 2.0 x 10-11 C2

Force F = Q1 Q2 / (4 pi epsilon0 r2) in N: F = 8.99 x 109 x 2.0 x 10-11 / 0.202 = 4.50 N, repulsive

Problem 3.

Product of the charge sizes in C2: Size of Q1 Q2 = 1.5 x 10-6 x 6.0 x 10-6 = 9.0 x 10-12 C2

Force F = Q1 Q2 / (4 pi epsilon0 r2) in N: F = 8.99 x 109 x 9.0 x 10-12 / 0.152 = 3.60 N, attractive

Check your understanding

A fixed positive source produces an outward field of 400 N/C at a point. Replace a small -2.00 nC test charge there with +2.00 nC, keeping the source unchanged. What happens?

  1. The field doubles because the test charge has changed sign.
  2. The field stays 400 N/C outward; the test-charge force becomes 8.00 x 10-7 N outward.
  3. The source field reverses, while the force on the test charge stays inward.
Show the answer

The field stays 400 N/C outward; the test-charge force becomes 8.00 x 10-7 N outward.

E is unchanged in the test-charge approximation. Multiplying by positive q gives a force along E, with magnitude (2.00 x 10-9)(400) = 8.00 x 10-7 N.

Check the permittivity units

The supplied permittivity is ε0 = 8.85 × 10-12 F/m. F here is the farad, defined in capacitance. Rearranging Coulomb's law gives:

Unit of ε0 = C2/(N m2) = F/m
= kg-1 m-3 s4 A2

This follows from C = A s and N = kg m/s2. K is the reciprocal 4πε0 factor. Separately rounded K and ε0 may produce slightly different final calculator digits.

Practise exam questions

Write your answer first, then tick the mark points it makes.

Exam-style question

Define electric field strength at a point. [2]

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Exam-style question

Calculate the electric field strength 0.10 m from a point charge of +2.0 nC in air. Take 1/(4πε0) = 8.99 x 109 N m2 C-2. [3]

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