What does "force per unit charge" mean physically?

It is the force a charge of one coulomb would feel at that point. A charge Q feels QE.

Why can field lines never cross?

At a crossing the field would have two directions at once, but a positive charge placed there can only be pushed one way.

Is the field between parallel plates truly uniform?

Almost exactly in the central region; it bends and weakens near the edges. Exam questions assume uniformity between the plates.

How is electric deflection like projectile motion?

Velocity perpendicular to the field is constant while a constant force acts along the field, producing a parabolic path — the same mathematics as a ball under gravity.

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7.1ElectricityCore

Electric Fields

A charged object affects the space around it so that any other charge placed there feels a force. That region of influence is an $electric field$ , and its strength at a point is the force it would exert on each coulomb of positive charge placed there.

40 min Topic 7 · Electric and Magnetic Fields

What you'll be able to do

Define electric field strength as force per unit charge
Use Coulomb’s law and $E = k Q / r^{2}$ for radial fields
Use $E = V / d$ for uniform fields between parallel plates
Describe and calculate the motion of charged particles in a uniform field

Field strength and field lines

Electric field strength at a point is the force per unit positive charge placed there: $E = F / Q$ , in $N C^{- 1}$ . It is a vector, pointing in the direction a positive charge would be pushed.

Field lines show direction: they start on positive charges and end on negative ones. Their spacing shows strength — closer lines mean a stronger field — and they never cross, since the field has one direction at each point.

A negative charge in a field experiences a force opposite to the field direction.

E = \frac{F}{Q}

Field strength, force per unit charge.

Tip — The field direction is defined for a $positive$ test charge. An electron accelerates against the field lines.

Coulomb’s law and radial fields

The force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation: $F = \frac{k Q _{1} Q _{2}}{r ^{2}}$ , with $k = \frac{1}{4 π ε _{0}} = 8.99 \times 1 0^{9} N m^{2} C^{- 2}$ .

A positive result means repulsion (like charges); negative means attraction.

Dividing by a test charge gives the field of a point charge, $E = k Q / r^{2}$ — a $radial$ field whose strength falls with the square of distance. A charged sphere behaves as if all its charge were at its centre, for points outside it.

F = \frac{k Q _{1} Q _{2}}{r ^{2}}

Coulomb’s law.

E = \frac{k Q}{r ^{2}}

Radial field of a point charge.

F = \frac{8.99 \times 1 0 ^{9} \times 3.0 \times 1 0 ^{- 9} \times 5.0 \times 1 0 ^{- 9}}{( 0.020 ) ^{2}}

= \frac{1.349 \times 1 0 ^{- 7}}{4.0 \times 1 0 ^{- 4}}

Answer

3.4 \times 1 0^{- 4} N

, attractive.

Uniform fields

Between two parallel charged plates, the field is $uniform$ — the same strength and direction everywhere, except near the edges. Field lines are parallel and equally spaced.

Its strength depends on the potential difference and plate separation: $E = V / d$ . This shows the equivalent unit $V m^{- 1}$ .

Moving the plates apart at constant voltage weakens the field; raising the voltage strengthens it.

E = \frac{V}{d}

Uniform field between parallel plates.

E = \frac{600}{4.0 \times 1 0 ^{- 3}} = 1.5 \times 1 0^{5} V m^{- 1}

F = e E = 1.60 \times 1 0^{- 19} \times 1.5 \times 1 0^{5}

Answer

E = 1.5 \times 1 0^{5} V m^{- 1}

;

F = 2.4 \times 1 0^{- 14} N

Tip — Convert plate separation from millimetres to metres. The field strength comes out a thousand times too small otherwise.

Charged particles in a uniform field

A charged particle in a uniform field feels a constant force $F = QE$ , so it has constant acceleration. Moving along the field, it simply speeds up; the energy gained crossing a p.d. $V$ is $Q V$ .

Entering $perpendicular$ to the field, it behaves like a projectile. Its velocity across the plates is unchanged, while it accelerates uniformly along the field direction, tracing a parabola.

The analysis uses the same method as projectile motion: find the time spent between the plates from the constant perpendicular velocity, then use suvat for the deflection.

a = \frac{QE}{m}

Constant acceleration.

\frac{1}{2} m v^{2} = Q V

Energy from accelerating through

V

1Time between plates:

t = 0.050/3.0 \times 1 0^{7} = 1.667 \times 1 0^{- 9}

a = e E / m = 1.60 \times 1 0^{- 19} \times 2.0 \times 1 0^{4} /9.11 \times 1 0^{- 31} = 3.51 \times 1 0^{15} m s^{- 2}

3Deflection:

\frac{1}{2} a t^{2} = 0.5 \times 3.51 \times 1 0^{15} \times (1.667 \times 1 0^{- 9})^{2}

Answer

4.9 \times 1 0^{- 3} m

, about 4.9 mm.

Equation recap

E = F / Q

Field strength.

F = k Q_{1} Q_{2} / r^{2}

Coulomb’s law.

E = k Q / r^{2}

Radial field.

E = V / d

Uniform field.

Common mistakes to avoid

Forgetting to square the separation in Coulomb’s law.

The force is inversely proportional to

r^{2}

Assuming field strength between plates varies with position.

Away from the edges the field is uniform:

E = V / d

everywhere.

Giving an electron a force along the field direction.

Negative charges experience force opposite to the field.

Including gravity when deflecting electrons.

Its effect is negligible compared with the electric force.

Key takeaways

Electric field strength is force per unit positive charge, $E = F / Q$ .
Coulomb’s law: $F = k Q_{1} Q_{2} / r^{2}$ ; a point charge has radial field $E = k Q / r^{2}$ .
Parallel plates give a uniform field $E = V / d$ .
Charges in a uniform field have constant acceleration $QE / m$ .
Perpendicular entry gives projectile-like parabolic deflection.

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Electric Fields

Field strength and field lines

Coulomb’s law and radial fields

Uniform fields

Charged particles in a uniform field

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Electric Fields

Field strength and field lines

Coulomb’s law and radial fields

Uniform fields

Charged particles in a uniform field

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