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:
$$F = qE$$
The 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.
- 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: $$a = \frac{3.2 \times 10^{-17} \, \text{N}}{9.11 \times 10^{-31} \, \text{kg}}$$ $$ = 3.5 \times 10^{13} \, \text{m s}^{-2}$$
Parabolic Path
- The vertical displacement $y$ of the particle over time $t$ is given by: $$y = \frac{1}{2} \frac{qE}{m} t^2$$
- Meanwhile, the horizontal displacement $x$ is: $$x = ut$$
Combining these equations shows that $y$ is proportional to $x^2$, confirming the parabolic nature of the path.
Work-Energy Relation: Kinetic Energy Gain
Work Done by the Electric Field
- When a charged particle moves through a potential difference $V$, the work done on it by the electric field is: $$W = qV$$
- This work is converted into the particle’s kinetic energy.
Kinetic Energy Gain
If the particle starts from rest, its gain in kinetic energy $\Delta E_K$ is equal to the work done:
$$\Delta E_K = \frac{1}{2} mv^2 = qV$$
A proton is accelerated from rest through a potential difference of $120 \, \text{V}$. Calculate the final velocity of the proton.
Solution
$$W = qV = (1.6 \times 10^{-19} \, \text{C})(120 \, \text{V}) $$
$$= 1.92 \times 10^{-17} \, \text{J}$$
This is equal to its kinetic energy:
$$\frac{1}{2} mv^2 = 1.92 \times 10^{-17} \, \text{J}$$
Solving for $v$:
$$v = \sqrt{\frac{2 \times 1.92 \times 10^{-17} \, \text{J}}{1.67 \times 10^{-27} \, \text{kg}}} $$
$$= 1.5 \times 10^5 \, \text{m s}^{-1}$$
Applications: Particle Accelerators and Electron Beams
Particle Accelerators
- Particle accelerators use electric fields to accelerate charged particles to high speeds.
- These particles are then used in experiments to study the fundamental structure of matter.
- The Large Hadron Collider (LHC) at CERN accelerates protons to nearly the speed of light using electric fields.
- These protons collide with each other, creating new particles that help scientists explore the origins of the universe.
Electron Beams
- Electron beams are used in technologies like cathode ray tubes (CRTs) in older televisions and oscilloscopes.
- In these devices, electrons are accelerated by electric fields and then deflected by magnetic fields to create images on a screen.
The principles of motion in electric fields are foundational for understanding modern technologies like X-ray machines, electron microscopes, and synchrotrons.
- What is the path of a charged particle entering a uniform electric field with an initial velocity perpendicular to the field?
- How does the work-energy principle explain the increase in kinetic energy of a charged particle moving through a potential difference?
- How are electric fields used in particle accelerators?
