Apply Coulomb's law F = k q1 q2 / r^2, the electric field of a point charge E = k Q / r^2, and the uniform electric field between parallel plates E = V / d to calculate forces, fields and the motion of charged particles

What this dot point is asking

QCAA wants you to apply Coulomb's law to the force between point charges, the inverse-square field around a point charge, and the uniform field between parallel plates, including the motion of charged particles accelerated or deflected by such fields. This dot point underpins IA1 short response on field-line diagrams and parallel-plate problems, and it is the foundation for Topic 2's later treatment of magnetic forces.

The answer

Coulomb's law

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

with Coulomb's constant $k = 8.99 \times 10^9$ N m$^2$ / C$^2$ (equivalently $1 / (4 \pi \varepsilon_0)$).

The force is along the line joining the charges, attractive if the charges have opposite signs and repulsive if they have the same sign. It is mutual (Newton's third law): both charges experience the same magnitude of force.

Coulomb's law mirrors Newton's law of universal gravitation structurally, but with sign and a far larger coupling constant.

Electric field of a point charge

The electric field $\vec{E}$ at a point is the force per unit positive test charge placed there:

Units: N/C or V/m (numerically equal). The field of a single point charge $Q$ at distance $r$ has magnitude:

The field points radially outward from a positive source charge and radially inward toward a negative source charge.

For two or more point charges, the net field is the vector sum of the individual fields. Standard QCAA stimulus shows two charges and asks for the field at a point on the perpendicular bisector or on the axis joining them.

Uniform field between parallel plates

Two parallel conducting plates separated by distance $d$ with a potential difference $V$ between them set up a uniform field in the region between the plates (away from the edges):

The field points from the positive plate to the negative plate. A positive charge accelerates with the field; a negative charge accelerates against it.

The work done by the field on a charge $q$ moving from one plate to the other:

If the charge is released from rest at one plate, all of this work becomes kinetic energy:

so the speed reached at the far plate is $v = \sqrt{2 q V / m}$.

Charged particle deflected sideways

A charged particle entering a parallel-plate region with a horizontal velocity $v_0$ perpendicular to the field experiences a force only in the field direction. The motion is exactly analogous to a horizontally launched projectile in gravity:

• Horizontal: constant velocity $v_0$, distance $x = v_0 t$.
• Vertical (along the field): acceleration $a = q E / m$, deflection $y = \tfrac{1}{2} a t^2$.

The deflection across a plate of length $L$ is $y = q E L^2 / (2 m v_0^2)$. After leaving the plates the particle continues in a straight line at the final exit velocity.

This is the operating principle of the cathode-ray oscilloscope and the deflecting plates in early TVs.

Try it: Electric field calculator. Compute the field of a point charge, the force on a test charge, or the field between parallel plates.

Field-line diagrams

QCAA frequently asks for field-line sketches. Conventions:

• Lines start at positive charges and end at negative charges (or extend to infinity).
• The tangent to a field line at any point gives the direction of $\vec{E}$ there.
• The density of field lines is proportional to the field strength.
• Field lines never cross.
• Inside a conductor in electrostatic equilibrium, $\vec{E} = 0$ and field lines start and end on the surface, perpendicular to it.

Standard sketches: a single positive charge (radial outward), a single negative charge (radial inward), a positive-negative pair (dipole), and parallel plates (uniform parallel lines between the plates, fringing near the edges).

How this appears in IA1 and IA2

IA1 data test. Expect a parallel-plate problem with a charged particle accelerated or deflected, asked for the final speed or deflection. Alternatively, a two-charge geometry with a question on the net field at a point. Markers focus on candidates who forget the direction of $\vec{E}$, treat the field of a negative charge as positive, or confuse the field direction with the force direction on a negative test charge.

IA2 student experiment. A field-only IA2 is rare in QCE Physics (typically Topic 2 IA2s use induction or transformers), but design discussions sometimes reference a Millikan-style charged-droplet apparatus or a parallel-plate beam-deflection demo. The theory section then uses $E = V/d$ and $qV = \tfrac{1}{2} m v^2$ to predict speeds and deflections.

Examples in context

Example 1. Gladstone industrial transformer insulation testing applies $V = 30 \text{ kV}$ across a $5.0 \text{ mm}$ oil-paper gap, giving $E = V/d = 6.0 \times 10^6 \text{ V m}^{-1}$. Breakdown strength of fresh transformer oil is around $15 \text{ MV m}^{-1}$, leaving a safety factor of $2.5$. The QCAA Unit 3 dot-point parallel-plate relation $E = V/d$ is what the design engineer applies.

Example 2. A cathode-ray-tube-style electron beam in a school physics demonstration set at $1500 \text{ V}$ over a $5.0 \text{ cm}$ accelerating gap gives $E = 3.0 \times 10^4 \text{ V m}^{-1}$. Force on the electron is $F = qE = 4.8 \times 10^{-15} \text{ N}$, acceleration $a = F/m_e = 5.27 \times 10^{15} \text{ m s}^{-2}$. The deflection-plate stage then operates exactly like projectile motion, an analogy QCAA EA Unit 3 thematic stems exploit.

Try this

Q1. State the relationship between the electric field and the potential difference between parallel plates. [2 marks]

• Cue. $E = V/d$ for uniform field.

Q2. Two point charges $+2.0 \text{ nC}$ and $-4.0 \text{ nC}$ are $0.10 \text{ m}$ apart. Calculate the force between them. ($k = 8.99 \times 10^9$). [3 marks]

• Cue. $F = kq_1q_2/r^2 = 8.99 \times 10^9 \times 2 \times 10^{-9} \times 4 \times 10^{-9} / 0.01 = 7.19 \times 10^{-6} \text{ N}$, attractive.

Q3. An electron enters a parallel-plate gap horizontally at $2.0 \times 10^7 \text{ m s}^{-1}$ with plate length $0.08 \text{ m}$ and gap $V = 200 \text{ V}, d = 0.02 \text{ m}$. (a) Calculate the field strength and the vertical acceleration. (b) Determine the transit time and vertical deflection. (c) Explain how the QCAA IA1 data test treats projectile-like motion of charges. [3+3+2 marks; ISMG: Analysis and interpretation]

• Cue. (a) $E = 10^4 \text{ V m}^{-1}$, $a = eE/m_e = 1.76 \times 10^{15} \text{ m s}^{-2}$; (b) $t = 4.0 \times 10^{-9} \text{ s}$, $y = \tfrac{1}{2}at^2 = 1.4 \times 10^{-2} \text{ m}$; (c) decompose horizontal at constant $v$, vertical with uniform $a$.