Electric fields, Coulomb's law and parallel plates: VCE Physics Unit 3

Syllabus dot point

“Describe electric fields using the field model, apply Coulomb's law $F = k q_1 q_2 / r^2$ and the relationships $E = F/q$ , $E = kQ/r^2$ for point charges and $E = V/d$ for the uniform field between parallel plates; identify the directions of field, force and acceleration of charged particles in uniform and radial fields”

11 min readUpdated 21 May 2026

Quick answer

An electric field has strength $E = kQ/r^2$ around a point charge and $E = V/d$ in the uniform region between parallel plates; a charge $q$ in the field feels force $F = qE$ that is along the field for positive charges and opposite for negative charges.

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What this dot point is asking
The answer
Examples in context
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Exam-style questions

What this dot point is asking

VCAA wants you to describe the electric field using the field model, apply Coulomb's law for point charges, calculate field strength for a point charge ( $E = kQ/r^2$ ) and between parallel plates ( $E = V/d$ ), and predict the direction of force and acceleration on a charged particle.

The answer

The field model

An electric field surrounds every electric charge. Another charge placed in the field experiences a force. The field is drawn with field lines that point in the direction of the force on a positive test charge.

Field lines start on positive charges and end on negative charges (or extend to infinity).
The density of field lines indicates field strength.
Between two parallel oppositely charged plates, field lines run from positive to negative, parallel and evenly spaced (uniform field).

Coulomb's law

The electric force between two point charges $q_1$ and $q_2$ separated by distance $r$ is:

F = \frac{k q_1 q_2}{r^2}

where $k = 9.0 \times 10^9$ N m squared per C squared. Like signs repel; opposite signs attract.

Electric field strength

The electric field strength $E$ at a point is the force per unit positive test charge:

E = \frac{F}{q}

Units are N/C, equivalent to V/m.

Point charge. A single point charge $Q$ produces a radial field:

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

pointing outward from a positive charge, inward toward a negative charge.

Parallel plates. Two large, parallel, oppositely charged plates separated by distance $d$ with potential difference $V$ produce a uniform field in the gap:

E = \frac{V}{d}

The field points from the positive plate to the negative plate.

Force and acceleration on a charged particle

A charge $q$ in field $E$ feels force $F = qE$ . By Newton's second law its acceleration is:

a = \frac{qE}{m}

Direction. A positive charge accelerates along the field. A negative charge accelerates opposite to the field.

In a uniform field between parallel plates, a charged particle entering perpendicular to the field follows a parabolic path (analogous to projectile motion with $g$ replaced by $a = qE/m$ ).

Comparing electric and gravitational fields

	Gravitational	Electric
Source	Mass	Charge
Force law	$F = G m_1 m_2 / r^2$	$F = k q_1 q_2 / r^2$
Field strength	$g = GM/r^2$	$E = kQ/r^2$
Direction of force	Always attractive	Repulsive (like) or attractive (unlike)
Test object	Mass	Charge

The two laws have the same $1/r^2$ form. The key difference is that gravity is always attractive, while the electric force can be either attractive or repulsive.

Worked example

An electron is released from rest near the negative plate of a parallel-plate setup with $V = 200$ V and $d = 4.0$ mm. Find the acceleration of the electron and the speed it reaches at the positive plate. ( $m_e = 9.1 \times 10^{-31}$ kg, $e = 1.6 \times 10^{-19}$ C.)

Step 1: Find the electric field between the plates

For a uniform field between parallel plates the field depends only on the voltage and the separation, so convert $d$ to metres first.

E = \frac{V}{d} = \frac{200}{0.004} = 5.0 \times 10^4 \text{ V/m.}

Step 2: Find the force on the electron

The electron carries charge $e = 1.6 \times 10^{-19}$ C. Because it is negatively charged, the force is directed toward the positive plate, opposite to the field direction.

F = eE = 1.6 \times 10^{-19} \times 5.0 \times 10^4 = 8.0 \times 10^{-15} \text{ N (toward the positive plate).}

Step 3: Find the acceleration

Applying Newton's second law with the electron's mass gives the acceleration. This is enormously large compared with everyday accelerations because the electron's mass is so small.

a = \frac{F}{m} = \frac{8.0 \times 10^{-15}}{9.1 \times 10^{-31}} = 8.8 \times 10^{15} \text{ m/s}^2.

Step 4: Find the final speed

Using kinematics with $u = 0$ and the full plate separation as the distance travelled:

v = \sqrt{2ad} = \sqrt{2 \times 8.8 \times 10^{15} \times 0.004} = \sqrt{7.0 \times 10^{13}} = 8.4 \times 10^6 \text{ m/s.}

The same answer follows from energy conservation: $\frac{1}{2} m v^2 = eV$ , giving $v = \sqrt{2eV/m}$ .

Try it: Electric field calculator - enter charge and distance, or voltage and plate separation, and get $E$ and $F$ .

Common traps

Confusing the direction of force on a negative charge: The field points from positive to negative; the force on a negative charge is opposite to the field, so it accelerates toward the positive plate.
Mixing units of $E$: N/C and V/m are the same unit. Use whichever matches the rest of the question.
Forgetting to square the radius: $E = kQ/r^2$ , not $kQ/r$ .
Applying $E = V/d$ for a point charge: $E = V/d$ only works for a uniform field (between parallel plates or inside a long parallel-plate capacitor). Use $E = kQ/r^2$ for a point charge.
Using nano and milli without converting: $1$ nC $= 10^{-9}$ C; $1$ mm $= 10^{-3}$ m. Convert before substituting.

Examples in context

Example 1. Australian Synchrotron parallel-plate deflector at Clayton. The Australian Synchrotron uses parallel-plate deflectors to steer electron bunches. A pair of plates separated by $d = 30$ mm with potential difference $V = 600$ V produces a uniform field $E = V/d = 600/0.03 = 2.0 \times 10^4$ V m $^{-1}$ . Force on an electron is $F = eE = 1.6 \times 10^{-19} \times 2.0 \times 10^4 = 3.2 \times 10^{-15}$ N, producing acceleration $a = F/m_e = 3.51 \times 10^{15}$ m s $^{-2}$ . With electron transit time of $1.0 \times 10^{-9}$ s, deflection from straight-line path is $\tfrac{1}{2}at^2 = 1.76$ mm, enough to direct beams into different beamlines.

Example 2. Lightning strike on Eureka Tower lightning rod. A typical thundercloud holds a charge of $\pm 30$ C at $5$ km altitude. The electric field at ground level can reach $E = 3 \times 10^4$ V m $^{-1}$ during an impending strike (the air's dielectric breakdown threshold). Treating cloud as a point charge: $E = kQ/r^2$ gives at the cloud surface (radius $200$ m), $E = 9 \times 10^9 \times 30 / 4 \times 10^4 = 6.75 \times 10^6$ V m $^{-1}$ . Eureka Tower's roof-mounted lightning rod concentrates the field at its tip, encouraging streamer formation and channelling the strike harmlessly to ground via a copper bus.

Try this

Q1. Define electric field strength and state its SI unit. [2 marks]

Cue. Force per unit positive charge, $E = F/q$ , in N C $^{-1}$ or V m $^{-1}$ .

Q2. Two parallel plates separated by $5$ mm have a potential difference of $400$ V. Calculate (a) the field strength, and (b) the force on a $2.0$ $\mu$ C charge between them. [4 marks]

Cue. (a) $E = V/d = 8.0 \times 10^4$ V m $^{-1}$ . (b) $F = qE = 2 \times 10^{-6} \times 8 \times 10^4 = 0.16$ N.

Q3. Refer to the synchrotron deflector. (a) Calculate the field between plates at $V = 600$ V, $d = 30$ mm. (b) Determine the acceleration of an electron in this field. (c) Explain why parallel plates produce a uniform field. [2+3+2 marks]

Cue. (a) $2.0 \times 10^4$ V m $^{-1}$ . (b) $3.51 \times 10^{15}$ m s $^{-2}$ . (c) Charge spreads uniformly on plate surfaces, giving constant field perpendicular to plates between them.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice3 marks

Two parallel plates are separated by 2.0 cm and have a potential difference of 600 V across them. Calculate the electric field strength between the plates and the force on an electron in the field. (e = 1.6 x 10^-19 C.)

Show worked answer →

For a uniform field between parallel plates:

$E = \frac{V}{d} = \frac{600}{0.020} = 3.0 \times 10^4$ V/m (equivalent to N/C).

Force on the electron:

$F = qE = 1.6 \times 10^{-19} \times 3.0 \times 10^4 = 4.8 \times 10^{-15}$ N.

The force on the electron is directed toward the positive plate (opposite to the field direction, because the electron is negatively charged).

Markers reward correct unit conversion (cm to m), the equivalence of V/m and N/C, and the explicit direction of the force on a negative charge.

Practice4 marks

Two point charges,

q_1 = +3.0

nC and

q_2 = -4.0

nC, are placed 5.0 cm apart in vacuum. Calculate the magnitude of the force between them and describe what happens to the force if the separation is halved. (k = 9.0 x 10^9 N m^2 / C^2.)

Show worked answer →

Apply Coulomb's law:

$F = \frac{k |q_1 q_2|}{r^2} = \frac{9.0 \times 10^9 \times (3.0 \times 10^{-9}) \times (4.0 \times 10^{-9})}{(0.05)^2}$

Numerator: $9.0 \times 10^9 \times 1.2 \times 10^{-17} = 1.08 \times 10^{-7}$ .

Denominator: $(0.05)^2 = 2.5 \times 10^{-3}$ .

$F = \frac{1.08 \times 10^{-7}}{2.5 \times 10^{-3}} = 4.3 \times 10^{-5}$ N (attractive, because opposite signs).

If the separation is halved, $r$ becomes $0.025$ m. Force scales as $1/r^2$ , so it increases by a factor of $2^2 = 4$ , giving $1.7 \times 10^{-4}$ N.

Markers reward correct unit handling (nC, cm), the attractive direction, and the $1/r^2$ scaling argument.