Uniform Electric Fields Revision Notes | A Level AQA Physics
The electric field strength between parallel plates is E = V/d, where V is the potential difference and d the separation. Free notes and 11 questions answered.
Updated 12 Sept 2026 · Revise Right Team
Uniform Electric Fields is part of Electric fields (A-level only) in A Level AQA Physics.
6 min
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31
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A Level AQA
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What this lesson covers
The electric field strength between parallel plates is E = V/d, where V is the potential difference and d the separation. Free notes and 11 questions answered.
Derive the equation $E = V/d$ for the electric field strength in a uniform field from the work done moving a charge between parallel plates
Calculate electric field strength, potential difference, and plate separation using $E = V/d$
Describe and explain the parabolic trajectory of a charged particle entering a uniform electric field at right angles
Apply the analogy between projectile motion and charged-particle motion to analyse trajectories quantitatively
Key ideas to keep in view
Use this lesson to stay inside Electric fields (A-level only) while connecting the detail back to the wider Fields and their consequences (A-level only) topic.
Uniform electric field: A region in which the electric field strength has the same magnitude and direction at every point. Produced between two large, parallel, oppositely charged conducting plates. Field lines are equally spaced and parallel.
Electric field strength (E): The force per unit positive charge at a point in a field. In a uniform field, given by $E = V/d$, where $V$ is the potential difference between the plates and $d$ is their separation. Measured in volts per metre (V m$^{-1}$) or equivalently newtons per coulomb (N C$^{-1}$).
Potential difference (V): The work done per unit charge moving a positive test charge between two points in a field. Measured in volts (V), where 1 V = 1 J C$^{-1}$.
Parabolic trajectory: The curved path followed by a charged particle entering a uniform electric field at right angles to the field lines. Arises because the particle has constant velocity parallel to the plates and constant acceleration perpendicular to them — directly analogous to projectile motion under gravity.
Revision notes
Read the core explanations from Uniform Electric Fields before testing yourself from memory.
1. Why Uniform Fields Matter
Two parallel metal plates connected to a battery create one of the most useful and controllable electric fields in physics. Unlike the field around a point charge — which weakens with distance — the field between the plates is uniform: same strength, same direction, everywhere between them.
That uniformity is what makes parallel-plate systems so powerful. Cathode ray tubes, ink-jet printers, and particle accelerators all exploit the predictable, constant force that a uniform field exerts on a moving charge. Understanding this field quantitatively is the goal of this lesson.
2. The Structure of a Uniform Electric Field
Picture two large horizontal metal plates separated by a distance $d$. The top plate is connected to the positive terminal of a supply and the bottom plate to the negative terminal, so a potential difference $V$ exists across the gap. The electric field lines run straight from the positive plate to the negative plate — parallel, equally spaced, and perpendicular to both plates.
At the edges of the plates the field lines bow outward (edge effects), but in the central region the field is genuinely uniform. AQA questions assume you are working in this central region, so edge effects are ignored unless the question explicitly raises them.
The two units — V m$^{-1}$ and N C$^{-1}$ — are genuinely equivalent, not just approximately equal. You will see both in AQA mark schemes, and either is acceptable. The derivation below shows exactly why they are the same.
3. Deriving E = V/d from Work Done
Here is where the equation earns its place. Rather than accepting $E = V/d$ as a given formula, AQA expects you to be able to derive it — and the derivation is short, logical, and worth knowing cold.
Start with two things you already know: the definition of electric field strength ($E = F/Q$) and the definition of potential difference ($V = W/Q$, where $W$ is work done). Moving a charge $Q$ from one plate to the other through a distance $d$ against the electric force requires work.
Worked example: Deriving E = V/d from Work Done
Show that the electric field strength between two parallel plates is $E = V/d$, starting from the work done moving a charge $Q$ through potential difference $\Delta V$ across plate separation $d$.
$W = F \times d$ — Work done equals force multiplied by displacement in the direction of the force. Here the force $F$ is the electric force on the charge and $d$ is the plate separation — the full distance moved between the plates.
$W = Q \Delta V$ — By definition, potential difference $\Delta V = W/Q$, so rearranging gives $W = Q \Delta V$. This is the key link: the work done on charge $Q$ moving through $\Delta V$ equals $Q \Delta V$.
$F d = Q \Delta V$ — Both expressions equal the work done, so they must be equal to each other. This is the AQA specification equation $Fd = Q\Delta V$.
$F = \frac{Q \Delta V}{d}$ — Divide both sides by $d$ to isolate the electric force on the charge.
$E = \frac{F}{Q} = \frac{Q \Delta V}{d \times Q} = \frac{\Delta V}{d}$ — The definition of electric field strength is $E = F/Q$. Substituting the expression for $F$ and cancelling $Q$ gives the result. Writing $\Delta V$ as $V$ (the pd between the plates), we arrive at $E = V/d$.
$\frac{V}{m} = \frac{J\,C^{-1}}{m} = \frac{N\,m\,C^{-1}}{m} = N\,C^{-1}$ — Since 1 J = 1 N m, dividing by metres cancels one factor of metres, leaving N C$^{-1}$. This confirms V m$^{-1}$ and N C$^{-1}$ are identical units.
Answer: $E = \frac{V}{d}$ (or equivalently $Fd = Q\Delta V$), with $E$ in V m$^{-1}$ or N C$^{-1}$.
Notice that the derivation rests on two definitions — work done and potential difference — plus the assumption that the field is uniform (so the force on the charge is constant across the full distance $d$). If the field were non-uniform, the force would vary with position and you could not write $W = Fd$ directly.
4. Using E = V/d in Calculations
The equation $E = V/d$ connects three quantities. Given any two, you can find the third — and AQA regularly asks you to rearrange it. The worked example below models the full method expected in a 'Calculate' question.
Worked example: Calculating E, then the force on a charge
Two parallel plates are separated by $d = 4.0 \times 10^{-2}$ m and connected to a 600 V supply. A proton (charge $e = 1.6 \times 10^{-19}$ C) is placed between the plates. Calculate (a) the electric field strength between the plates and (b) the electric force on the proton. [4 marks]
$V = 600$ V, $\quad d = 4.0 \times 10^{-2}$ m, $\quad Q = 1.6 \times 10^{-19}$ C — List every given value with units before substituting. This earns the method mark and prevents unit errors later.
$E = \frac{V}{d}$ — State the formula explicitly — this is the first method mark in a 'Calculate' question.
$E = \frac{600}{4.0 \times 10^{-2}} = \frac{600}{0.040} = 15\,000$ V m$^{-1}$ — Dividing 600 by 0.040: $600 \div 0.040 = 15\,000$. Check: $0.040 \times 15\,000 = 600$ ✓. So $E = 1.5 \times 10^{4}$ V m$^{-1}$.
$F = EQ$ — From the definition $E = F/Q$, rearranging gives $F = EQ$. State this formula for the second method mark.
$F = 1.5 \times 10^{4} \times 1.6 \times 10^{-19}$
$= 1.5 \times 1.6 \times 10^{4+(-19)}$
$= 2.4 \times 10^{-15}$ N — $1.5 \times 1.6 = 2.4$ and $10^{4} \times 10^{-19} = 10^{-15}$. The force on the proton is $2.4 \times 10^{-15}$ N directed from the positive plate towards the negative plate.
Answer: (a) $E = 1.5 \times 10^{4}$ V m$^{-1}$; (b) $F = 2.4 \times 10^{-15}$ N
5. Trajectory of a Charged Particle Entering the Field at Right Angles
Now picture a charged particle — say an electron — fired horizontally into the gap between two vertical plates. The particle enters at right angles to the electric field lines. What happens next is one of the most elegant results in A-level physics, and it maps perfectly onto something you already know: projectile motion.
In projectile motion, a ball thrown horizontally has two independent components of motion: constant horizontal velocity (no horizontal force) and constant downward acceleration due to gravity. The combination produces a parabola. A charged particle between the plates follows exactly the same logic — replace gravity with the electric force.
Parallel to the plates (horizontal): no electric force acts, so the particle travels at constant velocity $v_x$.
Perpendicular to the plates (vertical): a constant electric force $F = EQ$ acts, producing a constant acceleration $a = EQ/m$.
The two motions are independent and combine to give a parabolic path.
The particle curves towards the oppositely charged plate: a positive charge curves towards the negative plate; a negative charge curves towards the positive plate.
The trajectory is parabolic for the same mathematical reason as projectile motion: one direction has displacement proportional to time ($x = v_x t$) and the perpendicular direction has displacement proportional to time squared ($y = \frac{1}{2}at^2$). Eliminating $t$ gives $y \propto x^2$ — the equation of a parabola.
6. Quantitative Analysis of the Trajectory
AQA can ask you to calculate the deflection of a particle or its velocity components as it exits the field. The method mirrors the SUVAT approach from mechanics — you apply the equations of motion separately in each direction.
The key equations for the two directions are listed below. Notice that the acceleration $a$ in the perpendicular direction comes directly from Newton's second law applied to the electric force: $F = EQ = ma$, so $a = EQ/m$.
Worked example: Calculating the deflection of an electron between parallel plates
An electron (mass $m = 9.11 \times 10^{-31}$ kg, charge $e = 1.6 \times 10^{-19}$ C) enters a uniform electric field at right angles with a horizontal speed of $v_x = 3.0 \times 10^{7}$ m s$^{-1}$. The plates are 0.12 m long and have a potential difference of 250 V across a separation of $d = 2.0 \times 10^{-2}$ m. Calculate the vertical deflection of the electron as it passes through the field. [5 marks]
$E = \frac{V}{d} = \frac{250}{2.0 \times 10^{-2}} = \frac{250}{0.020} = 12\,500$ V m$^{-1}$ — Check: $0.020 \times 12\,500 = 250$ ✓. So $E = 1.25 \times 10^{4}$ V m$^{-1}$.
$a = \frac{EQ}{m} = \frac{1.25 \times 10^{4} \times 1.6 \times 10^{-19}}{9.11 \times 10^{-31}}$
$= \frac{2.0 \times 10^{-15}}{9.11 \times 10^{-31}}$
$= 2.195 \times 10^{15}$ m s$^{-2}$ — Numerator: $1.25 \times 1.6 = 2.0$ and $10^{4} \times 10^{-19} = 10^{-15}$, giving $2.0 \times 10^{-15}$ N. Dividing by $9.11 \times 10^{-31}$: $2.0 \div 9.11 \approx 0.2195$ and $10^{-15} \div 10^{-31} = 10^{16}$, so $a \approx 2.195 \times 10^{15}$ m s$^{-2}$. Keep this unrounded for the next step.
$t = \frac{L}{v_x} = \frac{0.12}{3.0 \times 10^{7}} = 4.0 \times 10^{-9}$ s — $0.12 \div 3.0 = 0.040$ and $10^{0} \div 10^{7} = 10^{-7}$... more carefully: $0.12 \div (3.0 \times 10^{7}) = (0.12 \div 3.0) \times 10^{-7} = 0.040 \times 10^{-7} = 4.0 \times 10^{-9}$ s. This is the time the electron spends between the plates.
$y \approx 1.8 \times 10^{-2}$ m $= 1.8$ cm — The deflection (1.8 cm) is less than the plate separation (2.0 cm), so the electron exits the field without hitting a plate — physically reasonable. Round to 2 significant figures to match the data given.
Answer: Vertical deflection $y \approx 1.8 \times 10^{-2}$ m (1.8 cm), directed towards the positive plate.
The sanity check in Step 5 is worth doing in every trajectory question. If your calculated deflection exceeds half the plate separation, the particle would have struck a plate before exiting — a sign that something has gone wrong in the working.
7. Comparing the Uniform Field with Gravitational and Radial Fields
Earlier in the course you met gravitational fields and the radial electric field around a point charge. The uniform field between parallel plates is a distinct third case, and AQA sometimes asks you to compare them. The table below consolidates the key differences.
Property
Uniform electric field (parallel plates)
Radial electric field (point charge)
Field line pattern
Parallel, equally spaced, perpendicular to plates
Radiate outward (or inward) from the charge
Field strength with distance
Constant throughout the gap
Decreases with distance: $E = kQ/r^2$
Force on test charge
Constant in magnitude and direction
Varies in magnitude; always directed radially
Trajectory of entering particle
Parabolic (constant acceleration)
More complex curve (varying acceleration)
Key equation
$E = V/d$
$E = kQ/r^2$
8. Exam Technique and Common Mistakes
The three specification points in this lesson — the derivation of $E = V/d$, the formula itself, and the parabolic trajectory — each have characteristic traps that appear repeatedly in AQA mark schemes. The guidance below targets the mistakes that cost students marks most often.
Together, the three specification points in this lesson form a coherent story: the uniform field is defined by $E = V/d$ (derived from $Fd = Q\Delta V$), it exerts a constant force on any charge within it, and that constant force — acting perpendicular to an initial velocity — produces the parabolic trajectory that underpins technologies from oscilloscopes to inkjet printers.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-12.
What is the electric field strength between parallel plates?
The electric field strength between parallel plates is E = V/d, where V is the potential difference between the plates in volts and d is the plate separation in metres. The unit V m⁻¹ is identical to N C⁻¹. Plates 4.0 × 10⁻² m apart on a 600 V supply give E = 600 ÷ 0.040 = 1.5 × 10⁴ V m⁻¹.
How do you derive E = V/d?
The work done moving a charge Q between the plates is W = Fd, and also W = QΔV from the definition of potential difference. Equating gives Fd = QΔV, so F = QΔV/d. Substituting into E = F/Q and cancelling Q gives E = ΔV/d. The derivation assumes a uniform field, so the force is constant across d.
Why does a charged particle follow a parabola between parallel plates?
A charged particle entering a uniform field at right angles follows a parabola, just like a projectile thrown horizontally. Parallel to the plates there is no force, so it moves at constant velocity; perpendicular to them, a constant force F = EQ gives a constant acceleration a = EQ/m. Since x = vt and y = ½at², y ∝ x².
How do you calculate the deflection of an electron between parallel plates?
Electron deflection between plates comes from E = V/d, a = EQ/m, the time in the field t = L/v, then y = ½at². For an electron at 3.0 × 10⁷ m s⁻¹ passing 0.12 m plates with 250 V across 2.0 × 10⁻² m: E = 1.25 × 10⁴ V m⁻¹ and t = 4.0 × 10⁻⁹ s, so y ≈ 1.8 × 10⁻² m.
Why is V m⁻¹ the same unit as N C⁻¹?
V m⁻¹ and N C⁻¹ are identical units because 1 V = 1 J C⁻¹ and 1 J = 1 N m, so V m⁻¹ = J C⁻¹ m⁻¹ = N m C⁻¹ m⁻¹ = N C⁻¹. Both units are used for electric field strength, and either is acceptable in an answer.
How do you find the force on a charge between parallel plates?
The force on a charge between parallel plates is F = EQ with E = V/d. Plates 4.0 × 10⁻² m apart on a 600 V supply give E = 1.5 × 10⁴ V m⁻¹, so a proton of charge 1.6 × 10⁻¹⁹ C feels F = 2.4 × 10⁻¹⁵ N, directed from the positive plate towards the negative plate.
What are edge effects between parallel plates?
Edge effects are where the electric field lines bow outward at the edges of parallel plates, so the field there is not uniform. In the central region the field is genuinely uniform, and calculations assume a charge is in that region unless a question raises edge effects.
Which way does an electron bend in an electric field?
In a uniform electric field, a negative charge such as an electron curves towards the positive plate, while a positive charge curves towards the negative plate. The path is a parabola, because the electric force is constant and perpendicular to the initial velocity.
What are uniform electric fields used for?
Uniform electric fields between parallel plates are used in cathode ray tubes, ink-jet printers, oscilloscopes and particle accelerators. These all exploit the predictable, constant force that a uniform field exerts on a moving charge.
How do you check whether an electron hits the plate?
An electron would strike a plate if its calculated deflection exceeds half the plate separation before it leaves the field, so checking this is a useful sanity test. A deflection of 1.8 cm between plates 2.0 cm apart shows the electron exits the field without hitting a plate.
What assumption is made when deriving E = V/d?
Deriving E = V/d assumes the field is uniform, so the force on a charge is constant across the full plate separation d and the work done is simply W = Fd. In a non-uniform field the force would vary with position, and W = Fd would not apply directly.