108-111 - Electric Fields, Field Strength and Coulomb's Law

Field As A Force Region

Imagine moving a tiny positive bead near a charged metal sphere. Far away, the bead may barely feel a force. Closer in, the force becomes larger. The bead is not needed to create the field; it is a probe that reveals the field that the source charge has already set up.

Two charges have different roles:

The word small matters. A test charge should not be so large that it noticeably changes the arrangement of charges creating the field. In exam calculations, this is usually assumed unless the question says otherwise.

The direction convention is simple: the electric field points in the direction of the force on a positive test charge. A negative charge in the same field feels a force in the opposite direction.

Common misconception contrast:

For Edexcel, state what is meant by an electric field needs the force idea. A vague answer such as "an area around a charge" is not enough unless it says that a charged particle experiences a force there.

Field Strength From Force

Field strength tells you how much force the field gives to each coulomb of charge placed at a point. That is why the defining equation is:

The unit comes straight from the definition: force per charge, so $\mathrm{N/C}$, written $\mathrm{N\,C^{-1}}$.

Worked example:

A $+2.0\,\mathrm{nC}$ test charge at a point experiences a force of $6.4 \times 10^{-5}\,\mathrm{N}$ to the east. Calculate the electric field strength at that point.

Decision point 1: the question gives the force on a known test charge, so use the definition:

Decision point 2: convert nanocoulombs into coulombs:

Substitute:

Direction: the test charge is positive, so the field points in the same direction as the force. The field is $3.2 \times 10^{4}\,\mathrm{N\,C^{-1}}$ east.

Now change only the test charge sign. If a $-2.0\,\mathrm{nC}$ charge at the same point experienced a force of $6.4 \times 10^{-5}\,\mathrm{N}$ west, the field would still be east. The negative charge is forced opposite to the field direction.

For an Edexcel “calculate” item, the route should show the equation, the unit conversion, substitution, and a final answer with unit. If direction is asked for, treat it as physics information, not decoration.

Coulomb's Law Between Charges

When two small charged objects are separated by a distance $r$, each charge exerts a force on the other along the line joining their centres. Coulomb's law gives the size of that force:

In most A Level calculations, find the magnitude of the force using the sizes of the charges. Then use the signs to decide the direction:

• like charges repel
• unlike charges attract

Worked example:

Two small spheres have charges $+3.0\,\mathrm{nC}$ and $-5.0\,\mathrm{nC}$. Their centres are $4.0\,\mathrm{cm}$ apart. Calculate the force between them and state whether it is attractive or repulsive.

Decision point 1: this asks for force between two charges, so use Coulomb's law, not $E = F/Q$.

Decision point 2: convert everything into SI units:

Substitute the magnitudes:

The charges have opposite signs, so the force is attractive. Each sphere experiences the same size force, but in opposite directions along the line joining their centres.

Misconception contrast:

Point Charge Field

Coulomb's law gives the force between two charges. Field strength asks a slightly different question: what force would each coulomb of test charge feel at this point?

Start from a source charge $Q$ and a test charge $q$ at distance $r$:

Divide by the test charge:

So the electric field strength due to a point charge is:

The important change is that the test charge has disappeared from the equation. The field at a point is set by the source charge and the distance from it. A test charge placed there feels a force because the field already exists.

Worked example: