Motion of Charged Particles in Uniform Electric Fields
Consider a charged particle entering a region between two parallel plates with an electric field.
How does it move? Does it follow a straight line, curve, or something else?
Acceleration in Uniform Electric Fields
Force on a Charged Particle
In a uniform electric field $E$, a charged particle with charge $q$ experiences a constant force $F$ given by:
$$F = qE$$
TipThe direction of the force depends on the sign of the charge:
- Positive charges move in the direction of the electric field.
- Negative charges move in the opposite direction.
Acceleration and Motion
- According to Newton’s second law, the acceleration $a$ of the particle is: $$a = \frac{F}{m} = \frac{qE}{m}$$
- This acceleration is constant, meaning the particle’s motion is predictable.
Consider a proton in a uniform electric field of $500 \, \text{N C}^{-1}$. Calculate the force on the proton and its acceleration.
Solution
$$F = qE$$
$$ = (1.6 \times 10^{-19} \, \text{C})(500 \, \text{N C}^{-1}) $$
$$= 8.0 \times 10^{-17} \, \text{N}$$
If the proton’s mass is $1.67 \times 10^{-27} \, \text{kg}$, its acceleration is:
$$a = \frac{8.0 \times 10^{-17} \, \text{N}}{1.67 \times 10^{-27} \, \text{kg}}$$
$$ = 4.8 \times 10^{10} \, \text{m s}^{-2}$$
Deflection in Plates: Parabolic Trajectories
Motion Perpendicular to the Field
When a charged particle enters a uniform electric field with an initial velocity perpendicular to the field, its path becomes parabolic.
Analogy- Think of this like a projectile launched horizontally under gravity.
- The electric field acts like a gravitational field, causing the particle to accelerate in the direction of the field.
Components of Motion
The motion can be broken down into two components:
- Horizontal motion (along the initial velocity):
- Constant velocity $v_x = u$ (no force acts in this direction).
- Vertical motion (along the electric field):
- Uniform acceleration $a = \frac{qE}{m}$.
Suppose an electron enters a uniform electric field with an initial velocity of $2.0 \times 10^6 \, \text{m s}^{-1}$. The field strength is $200 \, \text{N C}^{-1}$, and the electron’s mass is $9.11 \times 10^{-31} \, \text{kg}$.
Calculate the horizontal component of velocity and acceleration of the electron.
Solution
- Horizontal motion: Velocity remains constant: $$v_x = 2.0 \times 10^6 \, \text{m s}^{-1}$$
- Vertical motion: Force on the electron: $$F = qE = (1.6 \times 10^{-19} \, \text{C})(200 \, \text{N C}^{-1})$$ $$ = 3.2 \times 10^{-17} \, \text{N}$$
- Acceleration: $$a = \frac{3.2 \times 10^{-17} \, \text{N}}{9.11 \times 10^{-31} \, \text{kg}}$$ $$ = 3.5 \times 10^{13} \, \text{m s}^{-2}$$
