A field on its own does nothing you can feel, it is potential, waiting. Drop a charge into it and the field springs into action, pushing or pulling with a force you can calculate from a single, beautifully simple equation: $F = qE$. The size of the push depends on the charge; the direction depends on its sign.
A uniform electric field points to the right. You place a positive charge in it.
Before reading: which way does the force on the positive charge point, and which way would the force on a negative charge point? Write your prediction.
Field strength times charge gives the force
A field tells you the force a single coulomb would feel; multiply by how many coulombs you actually have and you get the real force.
The field strength $E$ is defined as the force per unit positive charge, so the force on a charge $q$ placed in a field of strength $E$ is the NESA formula-sheet relationship:
$$F = qE$$
This tells you two things at once:
The force on a charge in an electric field is $F = qE$. Its magnitude is proportional to both $q$ and $E$; a positive charge is forced along the field, a negative charge opposite to the field.
Pause, copy the highlighted equation and direction rule into your book before moving on.
Did you get this? True or false: a negative charge placed in a uniform electric field feels a force in the same direction as the field.
Putting a charge and a field into $F = qE$
We just saw that $F = qE$ gives both size and direction. That raises a question: how do we get a number out of it for a real charge? This card answers it → substitute the charge and field into $F = qE$, keeping SI units.
Force problems are direct: identify the charge and the field, multiply, and reason out the direction from the sign of the charge.
Use charge in coulombs and field in N C$^{-1}$; the force comes out in newtons.
An electric field of strength $2.5 \times 10^{4}\ \text{N C}^{-1}$ points east. Find the magnitude and direction of the force on a charge of $-6.0 \times 10^{-6}\ \text{C}$ placed in the field.
To find the force on a charge, substitute the charge magnitude and field into $F = qE$ (answer in newtons), then set the direction from the sign of the charge: along $E$ if positive, opposite $E$ if negative.
Add the highlighted worked method to your notes before the check below.
Quick check: A charge of $+2.0 \times 10^{-6}\ \text{C}$ sits in a field of $3.0 \times 10^{5}\ \text{N C}^{-1}$. What is the magnitude of the force on it?
Pattern A, Calculate then reason
A uniform field of $4.0 \times 10^{4}\ \text{N C}^{-1}$ points upward. In your book: (1) calculate the force on a charge of $+3.0 \times 10^{-6}\ \text{C}$; (2) calculate the force on a charge of $-3.0 \times 10^{-6}\ \text{C}$; (3) state the direction of each force and explain in one sentence why they differ.
Two truths, one lie: three of these statements about $F = qE$ are correct. Which one is the lie?
Dot point: Solve problems involving charges in electric fields using F = qE
$F = qE$ and $E = \dfrac{kq}{r^2}$ are two halves of one picture
We just learned to find the force from a known field. That raises a question: what if the field itself comes from another point charge? This card answers it → combine $F = qE$ with $E = \dfrac{kq}{r^2}$ to find the force directly.
The two equations you now know chain together: one charge makes the field, and that field pushes the second charge.
Think of it as a two-step chain. A source charge $q_1$ creates a field at distance $r$ given by $E = \dfrac{kq_1}{r^2}$ (L04). A second charge $q_2$ placed there feels a force $F = q_2 E$ (this lesson). Substituting one into the other:
$$F = q_2 E = q_2 \cdot \frac{kq_1}{r^2} = \frac{kq_1 q_2}{r^2}$$
That final form is Coulomb's law, which you will meet formally next lesson. For now, the key idea is that $F = qE$ is the bridge: it converts a field into the actual force a charge experiences.
$F = qE$ links a field to the force it exerts. Combined with $E = \dfrac{kq}{r^2}$ for a point source, it gives the force between two charges, $F = \dfrac{kq_1 q_2}{r^2}$, the bridge to Coulomb's law.
Pause, copy the highlighted connection into your book before the check below.
Last check: Decide if the statement is true or false.
If you double the charge placed at a fixed point in an unchanged electric field, the force on it doubles.
Fill the blanks: Complete the reasoning.
In a field of $E = 300\ \text{N C}^{-1}$ a charge of $2.0\ \text{C}$ feels a force of N. If the charge were negative, the force would point to the field.
Dot point: Solve problems involving charges in electric fields using F = qE
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. A uniform electric field of strength $5.0 \times 10^{4}\ \text{N C}^{-1}$ points to the right. A charge of $-8.0 \times 10^{-6}\ \text{C}$ is placed in the field. Calculate the magnitude of the force on the charge and state its direction, justifying the direction you give.
1 mark: uses $F = qE$ · 1 mark: $F = 0.40\ \text{N}$ · 1 mark: direction is to the left · 1 mark: justifies left because the charge is negative (force opposite to field)
ApplyBand 5(3 marks) 2. A small charged sphere experiences an upward electric force of $0.096\ \text{N}$ when placed in a uniform electric field of strength $1.2 \times 10^{4}\ \text{N C}^{-1}$ that is directed downward. Calculate the magnitude of the charge on the sphere and determine its sign, justifying the sign you give.
1 mark: rearranges $F = qE$ to $q = F/E$ · 1 mark: $q = 8.0 \times 10^{-6}\ \text{C}$ · 1 mark: the charge is negative, because the force is opposite to the field direction
Dot point: Solve problems involving charges in electric fields using F = qE
AnalyseBand 5(3 marks) 3. Two charges, $+2.0\ \mu\text{C}$ and $-6.0\ \mu\text{C}$, are placed at the same point in a uniform field. Compare the forces they experience in both magnitude and direction, and explain your reasoning using $F = qE$.
1 mark: same field $E$ acts on both · 1 mark: the $-6.0\ \mu\text{C}$ charge feels three times the force in magnitude · 1 mark: the forces are in opposite directions because the signs are opposite
Dot point: Solve problems involving charges in electric fields using F = qE
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): Using $F = qE$ with the magnitude of the charge: $F = (8.0 \times 10^{-6})(5.0 \times 10^{4}) = 0.40\ \text{N}$. The direction of the force is to the left. This is because the charge is negative: a negative charge experiences a force opposite to the direction of the electric field, and since the field points to the right, the force on the charge points to the left.
Q2 (3 marks): Rearranging $F = qE$ gives $q = F/E = 0.096/(1.2 \times 10^{4}) = 8.0 \times 10^{-6}\ \text{C}$, so the magnitude of the charge is $8.0\ \mu\text{C}$. The sign follows from the directions. The field points downward, and the field direction is by definition the direction of the force on a positive charge. The force on this sphere is upward, that is opposite to the field, so the sphere cannot be positive. It must carry a negative charge of $8.0\ \mu\text{C}$.
Q3 (3 marks): Both charges sit at the same point, so both experience the same field strength $E$. In $F = qE$ the force is directly proportional to the magnitude of the charge, and $6.0\ \mu\text{C}$ is three times $2.0\ \mu\text{C}$, so the negative charge experiences three times the force of the positive one. The directions differ because the signs differ: the $+2.0\ \mu\text{C}$ charge is pushed along the field direction, while the $-6.0\ \mu\text{C}$ charge is pushed in exactly the opposite direction. So the forces are antiparallel with magnitudes in the ratio $1:3$.
You can now find the force on any charge in a known field with $F = qE$ and reason out its direction. Next lesson studies the force directly between two charges with Coulomb's law, $F = \dfrac{kq_1 q_2}{r^2}$.
↩ Back to focus-area hubAt the start you predicted the directions of the forces on a positive and a negative charge in a rightward field. The positive charge is pushed right (along $E$) and the negative charge is pushed left (opposite $E$), both governed by $F = qE$, where the sign of $q$ flips the direction.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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