Topic 10.4.114 min read
Motion of a Charged Particle in an Electric Field
A charge q in an electric field E⃗ experiences force F⃗=qE⃗ and, if free, acceleration a⃗=qE⃗/m. A positive charge accelerates along the field while a negative charge accelerates opposite to it. Motion along the electric force converts electric potential energy into kinetic energy; moving a charge against the force requires external work and stores potential energy.
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The main idea
An electric field pushes or pulls a charged particle. A positive particle moves with the field arrows, while a negative particle is pushed in the opposite direction.
- q
- particle charge (C)
- \vec E
- electric field (N·C⁻¹)
- m
- particle mass (kg)
Energy changes
- Moving freely with the electric force: speed and kinetic energy increase.
- Electric potential energy decreases by the same amount.
- Moving against the force: an external agent must do work.
- That external work is stored as electric potential energy.
Recommended for Class 11
Force on a charged particle
When a charge q is placed in an electric field E⃗, it experiences an electric force. The charge does not have to touch either plate or source charge.
- \vec F
- electric force on the particle (N)
- q
- signed particle charge (C)
- \vec E
- electric field intensity (N·C⁻¹)
| Particle charge | Direction of F⃗ |
|---|---|
| q>0 | Same direction as E⃗ |
| q<0 | Opposite direction to E⃗ |
Acceleration in a uniform field
Newton's second law connects the electric force with the particle's acceleration.
- \vec a
- particle acceleration (m·s⁻²)
- m
- particle mass (kg)
If E⃗ is uniform, both the magnitude and direction of qE⃗ are constant. The particle therefore undergoes constant acceleration while it remains in the field.
| Change | Effect on acceleration magnitude |a|=|q|Em |
|---|---|
| |q| increases | Acceleration increases in direct proportion |
| E increases | Acceleration increases in direct proportion |
| m increases | Acceleration decreases |
Positive charge released between parallel plates
Between oppositely charged parallel plates, E⃗ points from the positive plate toward the negative plate. A positive charge released from rest is pushed in that direction.
As it moves freely, the electric field does positive work. Its speed and kinetic energy increase while its electric potential energy decreases.
- \Delta K
- change in kinetic energy (J)
- \Delta U
- change in electric potential energy (J)
- \Delta V
- final potential minus initial potential (V)
Moving a positive charge against the field
To move a positive charge opposite to E⃗ at constant speed, an external force must oppose the electric force. For a slow move, its magnitude equals qE.
When to use: Use as a magnitude for slow motion against a uniform field, where there is no change in kinetic energy.
- d
- distance moved opposite to the electric force (m)
The external work increases the charge's electric potential energy. When the charge is released, it moves back along E⃗ and regains an equivalent amount of kinetic energy.
Connection with plate potential difference
For a uniform field of magnitude E between plates separated by distance d, the magnitude of the potential difference is |ΔV|=Ed.
This equation shows that the energy transferred depends on charge and potential difference. The chosen route does not matter in an electrostatic field; only the initial and final potentials matter.
Speed after acceleration
If a particle starts from rest and electric potential energy is converted entirely into kinetic energy, energy conservation gives:
- v
- final speed (m·s⁻¹)
When to use: Use for a positive particle released from rest and moving to lower potential without other energy losses.
The same result follows from constant-acceleration kinematics in a uniform field: v²=u²+2ad with a=qE/m and Ed=Vi−Vf.
Positive and negative particles compared
| Feature | Positive charge | Negative charge |
|---|---|---|
| Force direction | Along E⃗ | Opposite E⃗ |
| Free motion from rest | Toward lower V | Toward higher V |
| Potential-energy change during free acceleration | Decreases | Decreases |
| Kinetic-energy change during free acceleration | Increases | Increases |
Textbook process in words
- Place a positive charge +q in the uniform field between opposite plates.
- If released, it moves from the positive plate toward the negative plate.
- It gains kinetic energy as electric potential energy decreases.
- To move it against the field, apply an external force of magnitude qE for a slow move.
- The external work increases its potential energy.
- When released again, it returns along the field and gains an equivalent amount of kinetic energy.
Exam-ready method
- Start with F⃗=qE⃗ and identify the sign of q.
- Use a⃗=qE⃗/m in a uniform field.
- State the direction relative to E⃗.
- For energy, use ΔK=−ΔU=−qΔV.
- For parallel plates, use |ΔV|=Ed.
- For release from rest, use ½mv²=q(Vi−Vf) when the right side is positive.
- Explain whether the electric field or an external agent does the work.
Exam Focus
Exam Focus
Important points
- F⃗=qE⃗ and charge-sign direction.
- a⃗=qE⃗/m in a uniform field.
- Free positive-charge motion between plates.
- External work against the field.
- ΔK=−ΔU=−qΔV.
- Conversion between potential and kinetic energy.
Important formulas
- F=q E
- a=q Em
- Wext=qEd=q|Δ V|
- Δ K=-qΔ V
- 12mv2=q(Vi-Vf)
Common exam traps
- Negative charges accelerate opposite to E⃗.
- For a slow move against the field, Wext=+ΔU while Wfield=−ΔU.
Long question structure
Force relation → direction by charge sign → acceleration → free motion → work and potential energy → release and kinetic energy
Common Mistakes
Common Mistakes
- Mistake
Assuming a negative charge accelerates along the electric field.
- Mistake
Using F=qE as an unsigned equation without interpreting the sign of q.
- Mistake
Saying potential energy increases during free motion caused by the field.
- Mistake
Mixing work done by the field with work done by an external agent.
- Mistake
Using ½mv²=qΔV without choosing initial and final potentials consistently.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
A charged particle in an electric field experiences the force F⃗=qE⃗.
In a uniform field, the force is constant, so the particle has constant acceleration a⃗=qE⃗/m.
A positive charge accelerates along E⃗; a negative charge accelerates opposite to E⃗.
A freely released positive charge moves from the positive plate toward the negative plate.
A charged particle in an electric field experiences F⃗=qE⃗ and, in a uniform field, constant acceleration a⃗=qE⃗/m.
F=q E, a=q Em, Δ K=-qΔ V
- Definition
- A charged particle in an electric field experiences F⃗=qE⃗ and, in a uniform field, constant acceleration a⃗=qE⃗/m.
- Formula
- F=q E, a=q Em, Δ K=-qΔ V
- Example
- A positive charge released between plates accelerates toward the negative plate while electric potential energy becomes kinetic energy.
- Critical warning
- Reverse the force direction for a negative charge and keep field work separate from external work.
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Exam Practice
Important Question
Q1. Explain the motion of a positive charge in a uniform electric field.
Concept
Q2. How does the motion of a negative charge differ from that of a positive charge?
Energy
Q3. Explain what happens when a positive charge is moved against an electric field and then released.
Derivation
Q4. Derive the speed of a positive particle released from rest through a potential drop.
Reasoning
Q5. Why is the acceleration constant between ideal parallel plates?