Lift a ball and it stores gravitational potential energy; release it and that energy becomes motion. A charge in an electric field behaves the same way. Move it against the field and you do work, storing electric potential energy; let it go and the field does work back, turning that stored energy into kinetic energy. The bookkeeping equation is $W = qEd$.
You push a positive charge from the negative plate toward the positive plate, against the field, then let it go.
Before reading: what do you think happens to the energy you put in, where does it go after you release the charge? Write your prediction.
Where a charge sits in a field decides how much energy it stores
A charge in a field is like a ball on a hill, its stored energy depends on where it is, and moving it changes that stored energy.
The electric potential energy of a charge depends on its position in the field, exactly as a ball's gravitational potential energy depends on its height. For a positive charge in a uniform field:
A charge's electric potential energy depends on its position in the field. Moving a positive charge against the field does work on it and increases its potential energy (like lifting a ball); moving it with the field decreases its potential energy (like a ball falling).
Pause, copy the highlighted idea into your book before moving on.
Did you get this? True or false: moving a positive charge against the electric field increases its electric potential energy.
Force times distance, written for a charge in a field
We just saw that moving a charge changes its potential energy. That raises a question: how much energy, exactly? This card answers it → the work done is $W = qEd$.
Work is just force times distance, and since the force on the charge is $qE$, the work over a distance $d$ is $qEd$.
In a uniform field, the force on a charge is constant at $F = qE$. Work is force times the distance moved along the field, so:
$$W = Fd = qEd$$
This $W$ is the energy transferred, the change in the charge's electric potential energy as it moves a distance $d$ through the field. It is measured in joules (J).
A charge of $+2.0 \times 10^{-6}\ \text{C}$ is moved $0.030\ \text{m}$ through a uniform field of strength $4.0 \times 10^{4}\ \text{N C}^{-1}$. Find the work done.
The work done moving a charge a distance $d$ through a uniform field is $W = qEd$, measured in joules. This work equals the change in the charge's electric potential energy.
Add the highlighted equation and worked method to your notes before the check below.
Quick check: A charge of $3.0 \times 10^{-6}\ \text{C}$ moves $0.020\ \text{m}$ through a field of $5.0 \times 10^{4}\ \text{N C}^{-1}$. How much work is done ($W = qEd$)?
Pattern A, Calculate then reason
A charge of $+4.0 \times 10^{-6}\ \text{C}$ is moved $0.050\ \text{m}$ through a uniform field of $3.0 \times 10^{4}\ \text{N C}^{-1}$. In your book: (1) calculate the work done $W = qEd$; (2) state whether the charge's potential energy increases or decreases if it moves against the field; (3) explain in one sentence what kind of energy this work becomes if the charge is then released and allowed to move back with the field.
Two truths, one lie: three of these statements about electric potential energy and POSITION are correct. Which one is the lie?
Dot point: Explain how the electric potential energy of a charged particle changes due to its position in an electric field
Potential energy and kinetic energy trade places, with nothing lost
We just learned to calculate the work done on a charge. That raises a question: what happens to that energy as the charge moves? This card answers it → it converts between electric potential energy and kinetic energy, conserving the total.
As a charge moves through a field, energy simply changes form, potential becomes kinetic, or kinetic becomes potential, and the total stays the same.
Consider a positive charge released from rest near the positive plate. As the field pushes it toward the negative plate, it moves with the field, so its electric potential energy decreases and its kinetic energy increases by the same amount. By the law of conservation of energy:
$$\text{loss in electric PE} = \text{gain in KE} \qquad qEd = \tfrac{1}{2}mv^2$$
The reverse happens if you push the charge back against the field: the kinetic energy (work you supply) is stored again as electric potential energy. Either way, energy is conserved, exactly the same accounting as a ball swapping gravitational potential energy for kinetic energy as it falls and rises.
When a charge moves with the field, electric potential energy converts to kinetic energy ($qEd = \tfrac{1}{2}mv^2$); moving against the field stores kinetic energy back as potential energy. Total energy is conserved throughout.
Pause, copy the highlighted energy transformation into your book before the check below.
Last check: Decide if the statement is true or false.
As a positive charge accelerates with the field, the electric potential energy it loses equals the kinetic energy it gains.
Fill the blanks: Complete the work calculation in a uniform field.
For a charge $q$ moved a distance $d$ along a uniform field $E$, the work done is $W = q$ $d$. A charge of $2.0\ \text{C}$ moved $3.0\ \text{m}$ along a field of $5.0\ \text{N C}^{-1}$ has $W =$ J.
Dot point: Solve problems involving changes in electric potential energy in a uniform electric field using W = qEd
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.
UnderstandBand 5(4 marks) 1. A positive charge is released from rest next to the positive plate of a uniform field and moves to the negative plate. Account for the energy transformations that occur, and state how the work done relates to the kinetic energy gained.
1 mark: charge moves with the field, losing electric potential energy · 1 mark: this is converted into kinetic energy · 1 mark: refers to conservation of energy · 1 mark: states $W = qEd =$ KE gained ($= \tfrac{1}{2}mv^2$)
ExplainBand 4(3 marks) 2. Explain how the electric potential energy of a positive charge changes as its position in a uniform electric field changes. Refer to movement along the field, against the field, and perpendicular to the field.
1 mark: moving against the field increases the electric potential energy · 1 mark: moving along the field decreases it · 1 mark: moving perpendicular to the field leaves it unchanged, because there is no displacement along the force
Dot point: Explain how the electric potential energy of a charged particle changes due to its position in an electric field
ApplyBand 4(3 marks) 3. A charge of $4.0\ \mu\text{C}$ is moved $0.025\ \text{m}$ against a uniform field of strength $6.0 \times 10^{4}\ \text{N C}^{-1}$. Calculate the work done on the charge and state what happens to its electric potential energy.
1 mark: uses $W = qEd$ · 1 mark: $W = 6.0 \times 10^{-3}\ \text{J}$ · 1 mark: the potential energy increases by that amount because the charge is moved against the field
Dot point: Solve problems involving changes in electric potential energy in a uniform electric field using W = qEd
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): As the positive charge moves from the positive plate to the negative plate it travels with the field, so the field does work on it. Its electric potential energy decreases, and this energy is transformed into kinetic energy, so the charge speeds up. By the law of conservation of energy, the electric potential energy lost equals the kinetic energy gained. The work done by the field is $W = qEd$, and this work equals the kinetic energy gained: $W = qEd = \tfrac{1}{2}mv^2$.
Q2 (3 marks): Electric potential energy is energy of position, so it changes only when the charge moves along the direction of the electric force. If a positive charge is pushed against the field, an external agent must do work against the electric force, and that work is stored, so the electric potential energy increases; this is the direct analogue of lifting a mass against gravity. If the charge moves along the field, the field itself does the work, the stored energy is released, and the electric potential energy decreases. If the charge moves perpendicular to the field lines, there is no displacement in the direction of the force, so no work is done by or against the field and the electric potential energy is unchanged. Position matters, and only the component of the movement along the field changes the energy.
Q3 (3 marks): The work done is $W = qEd = (4.0 \times 10^{-6})(6.0 \times 10^{4})(0.025)$. The product $qE$ is the constant force, $0.24\ \text{N}$, and multiplying by the displacement $0.025\ \text{m}$ gives $W = 6.0 \times 10^{-3}\ \text{J}$. Because the charge is positive and it is moved against the field direction, an external agent has to push it, so work is done on the charge and its electric potential energy increases by $6.0 \times 10^{-3}\ \text{J}$.
You can now calculate the work done on a charge with $W = qEd$ and account for the energy transformations as it moves through a field. Next lesson links this work to voltage with $W = qV$.
↩ Back to focus-area hubAt the start you predicted where the energy goes when you push a charge against the field and release it. The work you do is stored as electric potential energy; when you release the charge, the field converts that stored energy into kinetic energy as the charge accelerates back, total energy conserved, just like a ball lifted and dropped.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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