Model qualitatively and quantitatively the electric field, including direction and shape, produced between parallel charged plates and the potential difference, using E = V/d

What this dot point is asking

NESA wants you to describe the shape and direction of the electric field between two parallel plates, state the relationship $E = V/d$ between the field strength, the applied potential difference and the plate separation, and explain why the field in the central region is uniform. Diagrams of field lines, with arrows from positive to negative and even spacing in the middle, are standard.

The answer

Field shape between parallel plates

Two flat, conducting plates held at different potentials produce a characteristic field pattern.

• In the central region the field lines are straight, parallel, and evenly spaced. The field has the same magnitude and the same direction everywhere in this region. This is the uniform electric field.
• Near the edges of the plates the field lines bow outward. This is called fringing or the edge effect. The field is weaker and non-uniform there.
• Outside the plates (above the top plate or below the bottom plate) the field is essentially zero, provided the plates are large compared with the gap.

The field always points from the positive plate to the negative plate. For a positive test charge placed between the plates, the electric force is in the same direction as $E$; for a negative test charge, it is opposite.

The relationship E = V/d

For a uniform field, the potential difference $V$ between two points separated by a distance $d$ along the field direction is:

This single equation does a lot of work in HSC problems. A few consequences:

• Doubling $V$ (with $d$ fixed) doubles the field strength.
• Halving $d$ (with $V$ fixed) doubles the field strength.
• The field strength is constant across the gap: $E$ has the same value 1 mm from the top plate, in the middle, or 1 mm from the bottom plate.

Units: V/m is identical to N/C. A field of $1000$ V/m exerts a force of $1.6 \times 10^{-16}$ N on a single electronic charge.

Why the field is independent of position in the gap

This often catches students out. The field is uniform because each plate, if it were infinite, would produce a uniform field of magnitude $\sigma / (2 \varepsilon_0)$ everywhere on either side (where $\sigma$ is the surface charge density). Between two oppositely charged plates the fields from both add; outside, they cancel. The result is a constant field in the gap that does not depend on how close you are to either plate.

You do not need to derive this for the HSC, but you should be able to state that the field is uniform and that $E = V/d$ everywhere in the central region.

Diagrams you should be able to draw

• Two long horizontal plates with $+$ on the top, $-$ on the bottom.
• Five or six straight, vertical, evenly spaced arrows pointing downward in the central region.
• At the left and right ends, field lines curving outward (fringing).
• No field lines above the top plate or below the bottom plate (or only very faint ones).
• A test charge placed somewhere in the middle, with a force arrow.

Markers love a clean labelled diagram. Reserve space for one even in a short answer.

Worked numerical examples

Example 1. A capacitor has plates 4.0 mm apart and stores a potential difference of 12 V. The field strength is $E = 12 / 0.0040 = 3.0 \times 10^3$ V/m.

Example 2. A 9.0 V battery is connected to two plates separated by 3.0 cm. What is the force on an electron in the gap?

$E = 9.0 / 0.030 = 300$ V/m.
$F = qE = 1.60 \times 10^{-19} \times 300 = 4.8 \times 10^{-17}$ N.

The force is constant everywhere in the gap and is directed from the negative plate toward the positive plate (electrons are pulled toward $+$).

Example 3. The breakdown field of dry air is about $3 \times 10^6$ V/m. The maximum voltage you can apply across a 1.0 mm gap before the air ionises and a spark jumps across is $V = E d = 3 \times 10^6 \times 0.001 = 3000$ V, or 3 kV.

Try it: Electric field calculator for converting between $V$, $d$ and $E$.

Examples in context

Example 1. Capacitor field in a Snowy 2.0 power-factor correction bank. Each capacitor in the bank has parallel plates separated by $d = 0.50 \text{ mm}$ and rated at $V = 415 \text{ V}$. Field strength between the plates is $E = V/d = 415 / 5.0 \times 10^{-4} = 8.30 \times 10^5 \text{ V/m}$. This is below the breakdown strength of polypropylene dielectric ($\sim 5 \times 10^7 \text{ V/m}$), giving a safety margin of $\sim 60\times$. The field is uniform in the central region between the plates, with weaker fringing fields curving outward at the edges (which is why capacitor manufacturers use guard rings to keep the high-field region purely interior).

Example 2. Photocopier corona charging at a Sydney CBD office. A corona wire at $+5500 \text{ V}$ sits $d = 1.5 \text{ cm}$ above the photoreceptor drum (held at ground). Average field strength $E = V/d = 5500 / 0.015 = 3.67 \times 10^5 \text{ V/m}$. This exceeds air's breakdown threshold of $\sim 3 \times 10^6 \text{ V/m}$ at the wire (where field is concentrated by the small radius), causing ionisation and a corona discharge. The ionised molecules drift to the drum and deposit a uniform positive charge on the photoreceptor, ready for the laser-discharge step. Office printers across the Sydney CBD use the same principle on a smaller scale.

Try this

Q1. Define the electric field between two parallel charged plates and write the equation linking field strength, voltage and separation. [2 marks]

• Cue. Force per unit positive test charge; $E = V/d$ for parallel plates.

Q2. Two parallel plates are separated by $4.0 \text{ mm}$ with a $600 \text{ V}$ potential difference. Calculate the field strength and the force on a $+2.0 \mu\text{C}$ charge placed between them. [3 marks]

• Cue. $E = 600 / 4.0 \times 10^{-3} = 1.5 \times 10^5 \text{ V/m}$; $F = qE = 2.0 \times 10^{-6} \times 1.5 \times 10^5 = 0.30 \text{ N}$.

Q3. A pair of parallel plates is connected to a $9.0 \text{ V}$ battery. The plates are then moved further apart from $1.0 \text{ mm}$ to $3.0 \text{ mm}$ while remaining connected. (a) Find the field strength before and after. (b) Explain what happens to the field if the plates are instead disconnected before being separated (assuming charge is conserved). (c) State the units of $E$ and show $\text{V/m} = \text{N/C}$. [2+2+2 marks]

• Cue. (a) $9000 \to 3000 \text{ V/m}$. (b) Charge fixed, $V = Q/C$ and $C = \epsilon_0 A / d$, so $V$ rises and $E = V/d$ stays constant. (c) $F = qE$, so $\text{N/C} = \text{V/m}$.