18.2 Uniform electric fields
- Syllabus
- 9702–2028–2029
- Topic
- 18.2
- Level
- A2
Learning objectives
18.2.1E = ∆V / ∆d to calculate the field strength of the uniform field between• recall and use E = ∆V / ∆d to calculate the field strength of the uniform field between charged parallel plates18.2.2The effect of a uniform electric field on the motion of charged particles• describe the effect of a uniform electric field on the motion of charged particles
A uniform electric field has strength E=∆V/∆d between equipotential planes
In a uniform field, field strength magnitude is E=∆V/∆d, where ∆V is potential difference across perpendicular separation ∆d.
Use metres and volts, and remember field direction points from higher potential toward lower potential for a positive test charge.
A 600 V difference across 0.020 m gives E=3.0×10⁴ N C⁻¹.
The relation is for a uniform field; in a point-charge field strength changes with distance.
A uniform electric field gives a charged particle constant acceleration
F=qE,accelerationmagnitudea=∣q∣E/m
| Charge | Force and acceleration direction |
|---|---|
| q>0 | along the electric field |
| q<0 | opposite the electric field |
While the particle is in a uniform field, force and acceleration are constant. The velocity component parallel to the force changes uniformly; a perpendicular velocity component remains constant if other forces are negligible.
| Region | Motion for entry perpendicular to E |
|---|---|
| between the plates | parabolic curved path toward the force direction |
| beyond the plates | no electric force, so a straight line tangent to the path at exit |
At the same E, acceleration depends on |q|/m. A helium nucleus has twice a proton's charge but about four times its mass, so its electric acceleration and perpendicular velocity gain are half as large for the same transit time.
Do not continue curving the trajectory after the particle leaves the field. Also compare charge-to-mass ratio, not charge alone; a negative particle curves opposite the field arrows.