IB Physics Topic D.3.1 Charged Particle Motion in…

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:

$$F = qE$$

Tip

The direction of the force depends on the sign of the charge:

For a positive charge, the force and acceleration are in the direction of the electric field.
For a negative charge, the force and acceleration are opposite to the direction of the electric field.

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.

Example question

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):
1. Constant velocity $v_x = u$ (no force acts in this direction).
Vertical motion (along the electric field):
1. Uniform acceleration $a = \frac{qE}{m}$.

Example question

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:

$$\Delta E_K = \frac{1}{2} mv^2 = qV$$

Example question

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.

Note

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.

Tip

The principles of motion in electric fields are foundational for understanding modern technologies like X-ray machines, electron microscopes, and synchrotrons.

Active recall

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?

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:

$$F = qE$$

Tip

The direction of the force depends on the sign of the charge:

For a positive charge, the force and acceleration are in the direction of the electric field.
For a negative charge, the force and acceleration are opposite to the direction of the electric field.

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.

Example question

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):
1. Constant velocity $v_x = u$ (no force acts in this direction).
Vertical motion (along the electric field):
1. Uniform acceleration $a = \frac{qE}{m}$.

Example question

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:

$$\Delta E_K = \frac{1}{2} mv^2 = qV$$

Example question

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.

Note

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.

Tip

The principles of motion in electric fields are foundational for understanding modern technologies like X-ray machines, electron microscopes, and synchrotrons.

Active recall

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?

IB Physics D.3.1 Charged Particle Motion in Electric Fields Notes

IB Physics D.3.1 Charged Particle Motion in Electric Fields Notes

Acceleration in Uniform Electric Fields

Force on a Charged Particle

Acceleration and Motion

Deflection in Plates: Parabolic Trajectories

Motion Perpendicular to the Field

Components of Motion

Parabolic Path

Work-Energy Relation: Kinetic Energy Gain

Work Done by the Electric Field

Kinetic Energy Gain

Applications: Particle Accelerators and Electron Beams

Particle Accelerators

Electron Beams

Acceleration in Uniform Electric Fields

Force on a Charged Particle

Acceleration and Motion

Deflection in Plates: Parabolic Trajectories

Motion Perpendicular to the Field

Components of Motion

Parabolic Path

Work-Energy Relation: Kinetic Energy Gain

Work Done by the Electric Field

Kinetic Energy Gain

Applications: Particle Accelerators and Electron Beams

Particle Accelerators

Electron Beams

Topic A - Space, time and motion

Topic B - The particulate nature of matter

Topic C - Wave behaviour

Topic D - Fields

Topic E - Nuclear and quantum physics

Topic A - Space, time and motion

Topic B - The particulate nature of matter

Topic C - Wave behaviour

Topic D - Fields

Topic E - Nuclear and quantum physics

IB Physics D.3.1 Charged Particle Motion in Electric Fields Notes

IB Physics D.3.1 Charged Particle Motion in Electric Fields Notes

Acceleration in Uniform Electric Fields

Force on a Charged Particle

Acceleration and Motion

Deflection in Plates: Parabolic Trajectories

Motion Perpendicular to the Field

Components of Motion

Parabolic Path

Work-Energy Relation: Kinetic Energy Gain

Work Done by the Electric Field

Kinetic Energy Gain

Applications: Particle Accelerators and Electron Beams

Particle Accelerators

Electron Beams

Acceleration in Uniform Electric Fields

Force on a Charged Particle

Acceleration and Motion

Deflection in Plates: Parabolic Trajectories

Motion Perpendicular to the Field

Components of Motion

Parabolic Path

Work-Energy Relation: Kinetic Energy Gain

Work Done by the Electric Field

Kinetic Energy Gain

Applications: Particle Accelerators and Electron Beams

Particle Accelerators

Electron Beams