Now we put a charge inside the uniform field and watch what happens. A constant field means a constant force, and a constant force means a constant acceleration, exactly like a projectile under gravity. This lesson is where all your field equations finally work together on a single moving particle.
A charged particle is released from rest next to the positive plate of a uniform field and moves toward the negative plate.
Before reading: does it move at constant speed, or does it speed up as it crosses the gap? Write your guess and a one-sentence reason.
A uniform field acts just like gravity on a projectile
Because the field between the plates never changes, the force on the charge never changes, so it accelerates at a steady rate, exactly like a falling object.
Inside the uniform field, a charge $q$ feels a constant force. Building the chain of equations you already know:
$$E = \frac{V}{d} \qquad F = qE \qquad a = \frac{F}{m}$$
Because the acceleration is constant, the motion is governed by the same equations of motion (SUVAT) you used for projectiles:
A charge in a uniform field experiences a constant force $F = qE$ and so a constant acceleration $a = \dfrac{qE}{m}$. The motion follows the standard equations of motion, exactly like a projectile under gravity.
Pause, copy the highlighted chain into your book before moving on.
Did you get this? True or false: a charge released from rest in a uniform electric field moves across the gap at constant speed.
From plate voltage to particle acceleration
We just saw the chain of equations. That raises a question: how do we run it end to end on a real particle? This card answers it → find $E$, then $F$, then $a$, in that order.
These problems are a relay: each equation hands its answer to the next one until you reach the acceleration.
The reliable method is: (1) find the field with $E = \dfrac{V}{d}$; (2) find the force with $F = qE$; (3) find the acceleration with $a = \dfrac{F}{m}$.
Two plates are $4.0\ \text{cm}$ apart with $200\ \text{V}$ across them. A proton ($q = 1.6 \times 10^{-19}\ \text{C}$, $m = 1.67 \times 10^{-27}\ \text{kg}$) sits in the field. Find the acceleration of the proton.
For a charged particle between plates, work in order: $E = \dfrac{V}{d}$, then $F = qE$, then $a = \dfrac{F}{m}$. Because the field is uniform the acceleration is constant, so the equations of motion then give speed and time.
Add the highlighted method to your notes before the check below.
Quick check: A proton and an electron are placed in the same uniform field. Which experiences the greater acceleration?
E = V/d sets the force; the charge sets its direction; the mass sets how far it bends
Use the simulation. Set $V = 40\ \text{V}$ and $d = 2.0\ \text{cm}$. What are the field strength and the force on an electron?
Use the simulation. True or false: a proton fired between the same plates bends toward the positive plate, the same way as an electron.
Pattern A, Calculate in steps
Two plates are $1.0\ \text{cm}$ apart with $50\ \text{V}$ across them. An electron ($q = 1.6 \times 10^{-19}\ \text{C}$, $m = 9.11 \times 10^{-31}\ \text{kg}$) is in the field. In your book: (1) find $E$; (2) find the force $F$ on the electron; (3) find its acceleration $a$; (4) state which way the force acts relative to the field, with a one-sentence reason.
Match the pairs: Match each equation to what it lets you find for a charge between parallel plates.
Dot point: Solve problems relating to charged particles in electric fields using F = qE, E = V/d, E = kq/r², W = qV
$E = \dfrac{V}{d}$ for plates, $E = \dfrac{kq}{r^2}$ for point charges
We just learned the relay method for plates. That raises a question: how do you know which field equation to start with? This card answers it → read the geometry, plates use $E = \dfrac{V}{d}$, a point charge uses $E = \dfrac{kq}{r^2}$.
The shape of the problem tells you which field equation to reach for; once you have $E$, the rest of the method is identical.
You now have two ways to find a field $E$, and the situation tells you which to use:
Once you have $E$ from the correct equation, the force always comes from $F = qE$, and from there the acceleration and motion follow. So the only decision is which field equation the geometry calls for; everything after that is the same relay.
Pick the field equation from the geometry: $E = \dfrac{V}{d}$ for parallel plates, $E = \dfrac{kq}{r^2}$ for a point charge. Then $F = qE$ gives the force in either case, and the constant-acceleration motion follows.
Pause, copy the highlighted decision rule into your book before the check below.
Last check: Decide if the statement is true or false.
To find the field between two parallel plates you use $E = \dfrac{V}{d}$, but for the field at a distance from a single point charge you use $E = \dfrac{kq}{r^2}$.
Odd one out: three of these are constant for a charged particle moving through a uniform field between plates. Which one is not?
Dot point: Solve problems relating to charged particles in electric fields using F = qE, E = V/d, E = kq/r², W = qV
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 6(5 marks) 1. Two parallel plates are $2.0\ \text{cm}$ apart with $400\ \text{V}$ across them. A proton ($q = 1.6 \times 10^{-19}\ \text{C}$, $m = 1.67 \times 10^{-27}\ \text{kg}$) is released from rest at the positive plate. Calculate (a) the field strength, (b) the force on the proton, and (c) its acceleration.
1 mark: $E = \dfrac{V}{d} = 2.0 \times 10^{4}\ \text{V m}^{-1}$ · 1 mark: uses $F = qE$ · 1 mark: $F = 3.2 \times 10^{-15}\ \text{N}$ · 1 mark: uses $a = \dfrac{F}{m}$ · 1 mark: $a \approx 1.9 \times 10^{12}\ \text{m s}^{-2}$
ApplyBand 5(4 marks) 2. An electron starts from rest and is accelerated across the full gap between two parallel plates with a potential difference of $250\ \text{V}$. Using $W = qV$, calculate the kinetic energy it gains and hence its final speed. Take $m_e = 9.1 \times 10^{-31}\ \text{kg}$.
1 mark: $W = qV = 4.0 \times 10^{-17}\ \text{J}$ · 1 mark: states that all the work becomes kinetic energy · 1 mark: uses $E_k = \tfrac{1}{2}mv^2$ rearranged to $v = \sqrt{2E_k/m}$ · 1 mark: $v = 9.4 \times 10^{6}\ \text{m s}^{-1}$
Dot point: Solve problems relating to charged particles in electric fields using F = qE, E = V/d, E = kq/r², W = qV
AnalyseBand 5(3 marks) 3. A proton and an electron are released from rest at the midpoint between two charged parallel plates. Compare the magnitudes of the forces on them and the magnitudes of their accelerations, and explain the difference.
1 mark: the forces are equal in magnitude because the charge magnitudes are equal · 1 mark: the accelerations differ greatly because the masses differ · 1 mark: the electron's acceleration is about 1800 times larger, and they move towards opposite plates
Dot point: Solve problems relating to charged particles in electric fields using F = qE, E = V/d, E = kq/r², W = qV
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 (5 marks): (a) The uniform field is $E = \dfrac{V}{d} = \dfrac{400}{0.020} = 2.0 \times 10^{4}\ \text{V m}^{-1}$. (b) The force on the proton is $F = qE = (1.6 \times 10^{-19})(2.0 \times 10^{4}) = 3.2 \times 10^{-15}\ \text{N}$, directed from the positive plate toward the negative plate. (c) The acceleration is $a = \dfrac{F}{m} = \dfrac{3.2 \times 10^{-15}}{1.67 \times 10^{-27}} \approx 1.9 \times 10^{12}\ \text{m s}^{-2}$. This acceleration is constant because the field, and therefore the force, is uniform across the gap.
Q2 (4 marks): The work done on the electron is $W = qV = (1.6 \times 10^{-19})(250) = 4.0 \times 10^{-17}\ \text{J}$. The electron starts from rest and no other force does work on it, so all of that work appears as kinetic energy: $E_k = 4.0 \times 10^{-17}\ \text{J}$. Rearranging $E_k = \tfrac{1}{2}mv^2$ gives $v = \sqrt{2E_k/m} = \sqrt{2(4.0 \times 10^{-17})/(9.1 \times 10^{-31})}$. The quantity under the root is $8.79 \times 10^{13}$, so $v = 9.4 \times 10^{6}\ \text{m s}^{-1}$. Note this energy route needs neither the plate separation nor the field strength, which is why it is often the quicker of the two methods.
Q3 (3 marks): Both particles carry charge of the same magnitude, $1.6 \times 10^{-19}\ \text{C}$, and both sit in the same uniform field, so $F = qE$ gives forces of equal magnitude. Their directions are opposite because their signs are opposite, so the proton accelerates towards the negative plate and the electron towards the positive plate. The accelerations are very different, because $a = F/m$ divides by mass. The proton's mass is about $1.67 \times 10^{-27}\ \text{kg}$ while the electron's is about $9.1 \times 10^{-31}\ \text{kg}$, roughly 1800 times smaller, so the electron's acceleration is roughly 1800 times greater. Equal forces do not mean equal accelerations.
You can now chain $E = \dfrac{V}{d}$, $F = qE$ and $a = \dfrac{F}{m}$ to analyse a charged particle in a uniform field, and choose the right field equation for the geometry. Next lesson looks at the energy side: the electric potential energy and work done on a charge.
↩ Back to focus-area hubAt the start you predicted whether the particle moves at constant speed or speeds up. The answer is it speeds up: a uniform field gives a constant force, so the particle accelerates uniformly across the gap. If you said constant speed, revisit Card 1, constant field means constant force, not constant velocity.
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 →