Physics/Chapter 10: Electrostatics
Motion of a Charged Particle in an Electric Field
Explanation level
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.
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Exam preparation
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.
Common mistakes
- Assuming a negative charge accelerates along the electric field.
- Using F=qE as an unsigned equation without interpreting the sign of q.
- Saying potential energy increases during free motion caused by the field.
- Mixing work done by the field with work done by an external agent.
- Using ½mv²=qΔV without choosing initial and final potentials consistently.
Quick revision
Revision Zone
Motion of a Charged Particle in an Electric Field
Sticky Notes — Quick Revision
Key Point 1
- A charged particle in an electric field experiences the force F⃗=qE⃗.
Key Point 2
- In a uniform field, the force is constant, so the particle has constant acceleration a⃗=qE⃗/m.
Key Point 3
- Work done by the electric field increases kinetic energy and decreases electric potential energy.
Key Point 4
- For a slow move through distance d opposite to the force, the required external work has magnitude Wext=qEd=q|ΔV|.
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- 8 concept questions
- Easy → Medium → Hard
- Explanation after each answer
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?
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