When a charged particle enters an electric field, it experiences a force parallel to the direction of the electric field lines. If the electric field is uniform, then the charged particle will undergo uniform acceleration and move like a projectile.
Equations
Description
Equation
Electrostatic force
F=EQ
Newton's second law
F=ma
Variable definitions
Quantity Name
Symbol
Derived Unit
SI BASE Units
force
F
N
kgms−2
electricfieldstrength
E
NC−1
kgms−3A−1
charge
Q
C
As
mass
m
kg
kg
acceleration
a
ms−2
ms−2
Projectile motion
Charged particles in an electric field experience a force. When a charged particle moves though a uniform electric field, it will follow a trajectory similar to that of a mass in a gravitational field.
The charge travelling through the electric field will experience a force, which is dependent on its charge and the electric field strength.
The force is given as:
F=EQ
The force acts parallel to the direction of the electric field lines. A positive charge will move in the direction of the field lines and a negative charge will move in the opposite direction to the field lines.
1
More massive, positively charged particle
2
Less massive, negatively charged particle
Horizontal motion
The component of velocity, which is perpendicular to the electric field, remains unchanged throughout electric field. This is because there are no forces acting perpendicular to the electric field. Therefore, there is no acceleration in the horizontal component of the particle.
Vertical motion
There will be a change of velocity in the vertical motion of the particle, as this is the direction of the force.
As the force is constant, the charged particle will accelerate at a constant rate in the direction of the force. The acceleration can be calculated using Newton's second law:
F=ma
This means that the equations of motion (suvat) can be used as the acceleration is uniform.
Example
An electron is initially travelling horizontally at 5.4×105ms−1before passing between two charged parallel plates. The electric field strength between the plates is 0.40NC−1 and the vertical distance between the path of the electron and the positive plate is 16mm. Calculate the time taken for the electron to reach the plate.
Note: You only need the vertical components to work out the answer to this question. uv and sv are the vertical components for initial velocity and displacement respectively.
Next, write down the equations needed and rearrange if necessary:
F=EQF=mav→av=mFsv=uvt+21avt2
Find force:
F=0.40×1.6×10−19F=6.4×10−20N
Use the value of force to obtain a value for vertical acceleration:
av=9.11×10−316.4×10−20av=7.02525...×1010ms−2
Then use value for vertical acceleration to obtain a value for the time:
Note: As the initial vertical velocity is 0ms−1, the first term (uvt) is 0.