Physics (9702) | Uniform electric fields

✨ Study Notes: Uniform Electric Fields (9702 A Level Physics)

Welcome! This chapter is your bridge between the basic concepts of electricity (like voltage and current from AS Level) and how fields truly work in three dimensions. Don't worry if electric fields seem abstract—we will use familiar analogies, especially gravity, to make this topic crystal clear!

We are focusing specifically on Uniform Electric Fields, which are the simplest type to calculate and visualize. Mastering this section is essential for understanding the motion of charged particles, which is key for topics like particle physics and cathode ray tubes.

1. Defining and Visualizing a Uniform Electric Field

1.1 What makes a field "Uniform"?

A field is uniform if the Electric Field Strength (\(E\)) has the same magnitude and same direction at every single point within that region.

• Analogy: Think about the gravitational field right near the Earth's surface. If you drop a ball (the test mass), the gravitational acceleration \(g\) is constant (about \(9.81 \text{ m s}^{-2}\)) and always points straight down. This is essentially a uniform gravitational field.

1.2 Creating a Uniform Field

In practice, a uniform electric field is created by placing two large, flat, parallel conducting plates close together and connecting them to a power source (a potential difference, \(V\)).

1.3 Representing Uniform Fields with Field Lines

Electric fields are represented by lines, called field lines or lines of force.

• Field lines always point away from positive (+) charges and towards negative (-) charges.
• In a uniform field (like between parallel plates):
  

  1. The lines must be parallel to each other.

  2. The lines must be equally spaced.

• Note: The field is uniform only in the central region; it curves slightly at the edges (known as fringe effects), but for A Level calculations, we ignore these edges.

2. Quantifying Uniform Electric Field Strength (E)

We already know from AS Level that Electric Field Strength, \(E\), is defined as the force per unit positive charge: \[F = qE\]

However, for a uniform field created by parallel plates, we can relate \(E\) directly to the potential difference (\(V\)) applied and the distance (\(d\)) between the plates.

2.1 Field Strength and Potential Difference

The field strength \(E\) is also known as the potential gradient. This means it tells you how quickly the electric potential changes over distance.

The syllabus requires you to recall and use the equation: \[E = \frac{\Delta V}{\Delta d}\]

Where:

• \(E\) is the Electric Field Strength (the uniform field between the plates).
• \(\Delta V\) (or simply \(V\)) is the potential difference (voltage) between the plates (in Volts, V).
• \(\Delta d\) (or simply \(d\)) is the separation distance between the plates (in metres, m).

2.2 Units of E

This equation gives us a new common unit for \(E\): Volts per metre (\(\text{V m}^{-1}\)).

Did you know? \(1 \text{ V m}^{-1}\) is exactly equivalent to \(1 \text{ N C}^{-1}\) (Newtons per Coulomb), which comes from the definition \(E=F/q\). Both units are acceptable for electric field strength.

2.3 Calculation Example (Step-by-Step)

A potential difference of \(1200 \text{ V}\) is applied across two parallel plates separated by a distance of \(4.0 \text{ cm}\). Calculate the uniform electric field strength \(E\) between the plates.

1. Convert units: The separation distance \(d\) must be in metres.
   \(d = 4.0 \text{ cm} = 0.040 \text{ m}\).
2. Identify \(V\): \(V = 1200 \text{ V}\).
3. Apply the formula: \[E = \frac{V}{d}\] \[E = \frac{1200 \text{ V}}{0.040 \text{ m}}\]
4. Calculate result: \[E = 30000 \text{ V m}^{-1} \text{ or } 3.0 \times 10^4 \text{ N C}^{-1}\]

3. Motion of Charged Particles in a Uniform Field

This is where the physics gets exciting! A uniform field applies a constant force to any charged particle within it, causing uniform acceleration.

3.1 Determining the Force and Acceleration

When a particle of charge \(q\) and mass \(m\) enters the field \(E\), it immediately experiences an electrical force, \(F\).

Step 1: Calculate the Force \(F\) \[F = qE\]

• If the charge \(q\) is positive (e.g., a proton), \(F\) is in the same direction as the field lines (\(E\)).
• If the charge \(q\) is negative (e.g., an electron), \(F\) is in the opposite direction to the field lines (\(E\)).

Step 2: Calculate the Acceleration \(a\)
Since the field is uniform, the force \(F\) is constant. According to Newton's Second Law (\(F=ma\)): \[a = \frac{F}{m} = \frac{qE}{m}\]

Since \(q\), \(E\), and \(m\) are all constants, the acceleration \(a\) is uniform (constant). This means we can use the familiar kinematic (SUVAT) equations from AS mechanics!

3.2 Describing the Motion

The effect of the field depends on the initial motion of the particle:

Case A: Particle Moving Parallel to the Field

• If an electron is shot straight toward the negative plate (opposite to \(E\)), the electric force accelerates it towards the positive plate.
• Motion is simply a straight line with uniform acceleration.

Case B: Particle Moving Perpendicular to the Field

This is the most common scenario, similar to projectile motion in a gravitational field.

Imagine an electron entering the field horizontally (initial velocity \(v_x\)), while the electric field exerts an upward force (vertical acceleration \(a_y\)).

1. Horizontal Motion (x-direction): Since there is no force component parallel to the initial velocity (ignoring air resistance), the horizontal velocity \(v_x\) remains constant.
2. Vertical Motion (y-direction): The electric force \(F=qE\) acts perpendicular to the initial velocity, causing uniform acceleration \(a_y\) in the vertical direction.
3. Result: The combination of constant horizontal velocity and uniform vertical acceleration causes the particle's path to be a parabola (a curved arc).

Wait, why is it like projectile motion?
In projectile motion, gravity causes a constant downward acceleration \(g\). Here, the electric field causes a constant acceleration \(a = qE/m\). The mathematical description of the parabolic path is identical!