A point charge makes a messy, spreading field. Put two flat charged plates facing each other and something far tidier appears: a field that is the same strength everywhere between them, with straight, evenly spaced lines. This "uniform field" is the workhorse of electronics, and one short equation, $E = \dfrac{V}{d}$, controls it.
Two parallel plates are connected to a battery, one plate positive and one negative.
Before reading: do you think the electric field is stronger near the plates and weaker in the middle, or the same everywhere between them? Write your guess and a one-sentence reason.
Two flat plates make a field that is the same everywhere
Facing the two plates together turns the spreading field of a point charge into a neat, even field that points straight across the gap.
When two parallel plates are given opposite charges (one connected to the positive terminal of a supply, one to the negative), the field between them is uniform. On a field-line diagram this looks like:
The field between two oppositely charged parallel plates is uniform: straight, parallel, evenly spaced field lines run from the positive plate to the negative plate. Even spacing means the field strength is the same everywhere between the plates (except slight fringing at the edges).
Pause, copy the highlighted description into your book before moving on.
Did you get this? True or false: the evenly spaced field lines between two parallel plates show that the field strength is the same at every point between them.
Voltage and plate spacing set the field strength
We just saw that the field is uniform. That raises a question: what is its actual strength, and what controls it? This card answers it → the field equals the voltage divided by the gap, $E = \dfrac{V}{d}$.
The field between the plates is just the voltage shared out across the gap, more volts or a smaller gap means a stronger field.
For two parallel plates with a potential difference $V$ across them and a separation $d$, the NESA formula-sheet equation for the uniform field strength is:
$$E = \frac{V}{d}$$
This shows the two factors that affect the field: it is directly proportional to the voltage $V$ and inversely proportional to the separation $d$. The units V m$^{-1}$ are equivalent to N C$^{-1}$.
Two parallel plates are $5.0\ \text{mm}$ apart with a potential difference of $120\ \text{V}$ across them. Find the electric field strength between the plates.
The uniform field between parallel plates is $E = \dfrac{V}{d}$: directly proportional to the voltage and inversely proportional to the plate separation. Its units, V m$^{-1}$, are the same as N C$^{-1}$.
Add the highlighted equation and worked method to your notes before the check below.
Quick check: The voltage across two parallel plates is kept fixed while the plates are moved twice as far apart. The field strength between them:
Pattern A, Draw then calculate
Two parallel plates are $2.0\ \text{cm}$ apart with $300\ \text{V}$ across them. In your book: (1) draw the field-line diagram, labelling the positive and negative plates and the field direction; (2) calculate the field strength $E$; (3) state in one sentence what happens to $E$ if the voltage is doubled.
Fill the blanks: Complete the field-line picture between the plates.
Between two charged parallel plates the field lines are straight, and evenly spaced, running from the positive plate to the plate.
Dot point: Use field line diagrams to describe the electric field between 2 charged parallel plates
Spreading the charge over flat plates evens out the field
We just learned to calculate the field from $V$ and $d$. That raises a question: why is the field uniform here when a point charge's field was not? This card answers it → because the charge is spread evenly over large flat surfaces.
A single point sends its field out in all directions, but two large flat sheets of charge produce a steady, straight-across field in the space between them.
On a point charge, all the charge is concentrated at one spot, so its field spreads radially and weakens with distance. On a parallel plate, the charge is spread evenly across a wide flat surface. Every part of the positive plate pushes a test charge toward the negative plate, and these contributions combine to give a field that is the same strength and direction throughout the gap. Only near the edges, where the plates run out, do the lines bulge outward, the fringing effect, so the field is uniform everywhere except right at the edges.
The parallel-plate field is uniform because charge is spread evenly over wide flat plates, so the combined field is constant in strength and direction across the gap. Only at the edges (fringing) does the field stop being uniform.
Pause, copy the highlighted explanation into your book before the check below.
Last check: Decide if the statement is true or false.
The field between parallel plates is uniform because the charge is spread evenly over the flat plates, but it curves slightly at the edges.
Two truths, one lie: three of these statements about the STRENGTH of the field between parallel plates are correct. Which one is the lie?
Dot point: Analyse the factors that affect the strength of the electric field between 2 charged parallel plates using E = V/d
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 5(4 marks) 1. Two parallel plates are $8.0\ \text{mm}$ apart with a potential difference of $240\ \text{V}$ across them. Calculate the electric field strength between the plates and explain how the field would change if the plate separation were increased to $16\ \text{mm}$ at the same voltage.
1 mark: converts $d$ to $8.0 \times 10^{-3}\ \text{m}$ · 1 mark: uses $E = \dfrac{V}{d}$ · 1 mark: $E = 3.0 \times 10^{4}\ \text{V m}^{-1}$ · 1 mark: doubling $d$ halves the field (inverse proportionality)
ExplainBand 5(3 marks) 2. Explain why the field between two large parallel plates is uniform, using the field-line picture, and describe how the field-line diagram differs near the edges of the plates.
1 mark: charge spreads evenly over the facing surfaces · 1 mark: field lines are straight, parallel and equally spaced, so strength and direction are the same everywhere · 1 mark: near the edges the lines bulge outwards and spread, so the field weakens and changes direction
Dot point: Use field line diagrams to describe the electric field between 2 charged parallel plates
AnalyseBand 5(3 marks) 3. A pair of parallel plates is connected to a fixed supply. Describe two distinct changes that would each double the electric field strength between them, and explain why simply doubling the area of the plates would not.
1 mark: double the supply voltage · 1 mark: halve the plate separation · 1 mark: explains that $E = V/d$ contains no area term, so plate area does not change the field strength
Dot point: Analyse the factors that affect the strength of the electric field between 2 charged parallel plates using E = V/d
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): Converting the separation, $d = 8.0\ \text{mm} = 8.0 \times 10^{-3}\ \text{m}$. The uniform field strength is $E = \dfrac{V}{d} = \dfrac{240}{8.0 \times 10^{-3}} = 3.0 \times 10^{4}\ \text{V m}^{-1}$, directed from the positive plate to the negative plate. Because $E = \dfrac{V}{d}$ makes the field inversely proportional to the separation, increasing $d$ to $16\ \text{mm}$ (doubling it) at the same voltage would halve the field strength to $1.5 \times 10^{4}\ \text{V m}^{-1}$.
Q2 (3 marks): Charge spreads itself evenly over the facing surfaces of the two plates, so every part of the region between them sees the same arrangement of charge. The field lines therefore run straight across from the positive plate to the negative plate, parallel to one another and equally spaced. Equal spacing means equal strength, and parallel lines mean the same direction, so the field is uniform: a test charge feels the same force wherever it is placed in the central region. Near the edges this breaks down. There is no more plate beyond the edge, so the lines bulge outwards into the surrounding space, curving and spreading apart. Wider spacing means a weaker field, and the curvature means the direction is no longer perpendicular to the plates. That is the fringing effect, and it is why calculations using $E = V/d$ are applied to the central region.
Q3 (3 marks): The field between the plates is $E = V/d$, so the strength depends on only two quantities. One way to double it is to double the potential difference across the plates while leaving the separation alone, since $E$ is directly proportional to $V$. A second, independent way is to halve the separation while leaving the voltage alone, since $E$ is inversely proportional to $d$; bringing the plates half as far apart doubles the field. Doubling the area of the plates does neither. The area does not appear in $E = V/d$ at all, so it has no effect on the field strength in the central region. Larger plates do make the uniform region wider and push the fringing further from the centre, which improves the approximation, but the field strength itself is unchanged.
You can now describe the uniform field between parallel plates, draw its field lines, and calculate its strength with $E = \dfrac{V}{d}$. Next lesson sends charged particles through this field, combining $F = qE$ and $E = \dfrac{V}{d}$.
↩ Back to focus-area hubAt the start you predicted whether the field varies between the plates. The answer is the field is uniform, the same strength everywhere, shown by the straight, evenly spaced field lines. If you thought it was stronger near the plates, revisit Card 1: even spacing means constant strength.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
Tell us what you would like help with. Register interest; support details will be confirmed before enrolment.
Register interest →