Uniform electric fields

1. Overview

A uniform electric field is a region of space where the electric field strength $E$ is constant in both magnitude and direction at every point. This physical environment is created by applying a potential difference across two parallel conducting plates separated by a small distance. Within this field, any charged particle experiences a constant electric force, resulting in a uniform acceleration (or deceleration) according to Newton’s Second Law. This predictability makes uniform fields essential for controlling the trajectories of electrons and ions in technologies such as mass spectrometers, particle accelerators, and oscilloscope deflection plates.

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Key Definitions

• Electric Field Strength ($E$): The force per unit positive charge acting on a stationary charge at a point in the field. It is a vector quantity.
• Uniform Electric Field: A field where the electric field strength is the same at all points, represented by parallel, equally spaced field lines.
• Potential Difference ($\Delta V$): The work done per unit charge in moving a positive test charge between two points in an electric field.
• Equipotential Surface: A surface (or line in 2D) where all points are at the same electric potential. In a uniform field, these are planes parallel to the conducting plates. No work is done when moving a charge along an equipotential surface.

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Content

3.1 Field Strength between Charged Parallel Plates

When two parallel metal plates are separated by a distance $d$ and connected to a DC power supply providing a potential difference $\Delta V$, a uniform electric field is established in the gap between them.

The Mathematical Relationship: The magnitude of the electric field strength $E$ is directly proportional to the potential difference and inversely proportional to the separation distance:

$$E = \frac{\Delta V}{d}$$

• $E$: Electric field strength, measured in volts per metre ($\text{V m}^{-1}$) or newtons per coulomb ($\text{N C}^{-1}$).
• $\Delta V$: Potential difference between the plates, measured in volts ($\text{V}$).
• $d$: Separation distance between the plates, measured in metres ($\text{m}$).

Field Characteristics:

1. Direction: Field lines always point from the positive plate (higher potential) to the negative plate (lower potential).
2. Visual Representation: In a diagram, the field is shown as straight, parallel lines with arrows. The equal spacing between lines indicates that the field strength is constant.
3. Edge Effects (Fringing): Near the edges of the plates, the field lines "bulge" outward and become less dense. In this region, the field is non-uniform and weaker. For most A-Level calculations, we assume the plates are infinitely large or focus only on the region strictly between them where the field is uniform.
4. Potential Gradient: The electric field strength is equal to the negative of the potential gradient ($E = -\frac{\Delta V}{\Delta x}$). This means the potential decreases linearly as you move in the direction of the field lines.

Derivation of $E = \Delta V / d$:

1. Consider a charge $+q$ moved from the negative plate to the positive plate.
2. The work done ($W$) by an external force against the electric field is: $W = q\Delta V$.
3. Work done is also defined as $\text{Force} \times \text{distance}$ moved in the direction of the force: $W = F \times d$.
4. The electric force acting on the charge is $F = qE$.
5. Substituting force into the work equation: $W = (qE) \times d$.
6. Equating the two expressions for work: $q\Delta V = qEd$.
7. Dividing both sides by $q$ yields: $\Delta V = Ed$, which rearranges to $E = \frac{\Delta V}{d}$.

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3.2 Effect of a Uniform Electric Field on the Motion of Charged Particles

When a particle with charge $q$ and mass $m$ enters a uniform electric field, it experiences a constant force $F = qE$. According to $F = ma$, the particle will undergo a constant acceleration:

$$a = \frac{qE}{m} = \frac{q\Delta V}{md}$$

The resulting motion depends on the particle's initial velocity relative to the field lines.

Case A: Motion Parallel to Field Lines If a particle is released from rest or enters the field moving parallel to the field lines:

1. Force and Acceleration: The force acts in the same (or opposite) direction as the motion.
2. Trajectory: The particle moves in a straight line.
3. Kinematics: Because acceleration is constant, the equations of motion (SUVAT) apply:
   • $v = u + at$
   • $v^2 = u^2 + 2as$
   • $s = ut + \frac{1}{2}at^2$
4. Energy Transfer: The work done by the field on the particle results in a change in kinetic energy: $q\Delta V = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

Case B: Motion Perpendicular to Field Lines (Parabolic Path) If a particle enters the field with an initial velocity $v_x$ perpendicular to the field lines (e.g., an electron moving horizontally into a vertical field):

1. Horizontal Component ($x$-axis):
   • There is no force acting horizontally ($F_x = 0$).
   • The horizontal velocity $v_x$ remains constant.
   • Horizontal displacement: $x = v_x t$.
2. Vertical Component ($y$-axis):
   • A constant vertical force $F_y = qE$ acts on the particle.
   • This produces a constant vertical acceleration $a_y = \frac{qE}{m}$.
   • Initial vertical velocity $u_y = 0$.
   • Vertical displacement: $y = \frac{1}{2}a_y t^2 = \frac{1}{2}\left(\frac{qE}{m}\right)t^2$.
3. The Trajectory:
   • By substituting $t = \frac{x}{v_x}$ into the vertical displacement equation: $y = \left(\frac{qE}{2mv_x^2}\right)x^2$
   • Since $q, E, m,$ and $v_x$ are constants, $y \propto x^2$.
   • The path is a parabola, mathematically identical to a projectile in a uniform gravitational field.

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3.3 Worked Example 1 — Field Strength and Force

Two horizontal parallel plates are separated by a distance of $15.0\text{ mm}$. The upper plate is held at a potential of $+600\text{ V}$ and the lower plate is earthed ($0\text{ V}$). An alpha particle (charge $+2e$, mass $6.64 \times 10^{-27}\text{ kg}$) is placed between the plates.

Calculate:

1. The electric field strength between the plates.
2. The electric force acting on the alpha particle.

Solution: Step 1: Calculate Electric Field Strength ($E$)

• $\Delta V = 600 - 0 = 600\text{ V}$
• $d = 15.0\text{ mm} = 15.0 \times 10^{-3}\text{ m}$
• $E = \frac{\Delta V}{d}$
• $E = \frac{600}{15.0 \times 10^{-3}}$
• $E = 4.00 \times 10^4\text{ V m}^{-1}$ (directed downwards)

Step 2: Calculate Electric Force ($F$)

• $q = 2e = 2 \times (1.60 \times 10^{-19}\text{ C}) = 3.20 \times 10^{-19}\text{ C}$
• $F = qE$
• $F = (3.20 \times 10^{-19}) \times (4.00 \times 10^4)$
• $F = 1.28 \times 10^{-14}\text{ N}$ (directed downwards, as the particle is positive)

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3.4 Worked Example 2 — Deflection of an Electron

An electron is accelerated to a horizontal velocity of $2.0 \times 10^7\text{ m s}^{-1}$. It enters the space between two parallel plates of length $4.0\text{ cm}$. The electric field strength between the plates is $5000\text{ V m}^{-1}$ and is directed vertically upwards.

Calculate the vertical deflection of the electron as it leaves the plates. (Mass of electron $m_e = 9.11 \times 10^{-31}\text{ kg}$; Charge $e = 1.60 \times 10^{-19}\text{ C}$)

Solution: Step 1: Determine the time ($t$) spent in the field The horizontal velocity is constant.

• $x = 4.0\text{ cm} = 0.040\text{ m}$
• $v_x = 2.0 \times 10^7\text{ m s}^{-1}$
• $t = \frac{x}{v_x} = \frac{0.040}{2.0 \times 10^7} = 2.0 \times 10^{-9}\text{ s}$

Step 2: Determine the vertical acceleration ($a_y$) The force on the electron is downwards (opposite to the upward field).

• $a_y = \frac{eE}{m}$
• $a_y = \frac{(1.60 \times 10^{-19}) \times 5000}{9.11 \times 10^{-31}}$
• $a_y = 8.78 \times 10^{14}\text{ m s}^{-2}$

Step 3: Calculate vertical displacement ($y$)

• $y = u_y t + \frac{1}{2}a_y t^2$ (where $u_y = 0$)
• $y = 0.5 \times (8.78 \times 10^{14}) \times (2.0 \times 10^{-9})^2$
• $y = 0.5 \times (8.78 \times 10^{14}) \times (4.0 \times 10^{-18})$
• $y = 1.756 \times 10^{-3}\text{ m}$
• $y \approx 1.8\text{ mm}$ (to 2 s.f.)

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Key Equations

Equation	Symbols	SI Units	Status
$E = \frac{\Delta V}{d}$	$E$: Field strength, $V$: Potential, $d$: Separation	$\text{V m}^{-1}$, $\text{V}$, $\text{m}$	Memorize
$E = \frac{F}{q}$	$F$: Force, $q$: Charge	$\text{N C}^{-1}$, $\text{N}$, $\text{C}$	Data Sheet
$F = qE$	$F$: Force, $q$: Charge, $E$: Field strength	$\text{N}$, $\text{C}$, $\text{V m}^{-1}$	Memorize
$W = q\Delta V$	$W$: Work, $q$: Charge, $V$: Potential	$\text{J}$, $\text{C}$, $\text{V}$	Data Sheet
$a = \frac{qE}{m}$	$a$: Acceleration, $m$: Mass	$\text{m s}^{-2}$, $\text{kg}$	Memorize
$\Delta E_k = q\Delta V$	$E_k$: Kinetic Energy	$\text{J}$	Memorize

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Common Mistakes to Avoid

• ❌ Confusing $d$ with $r$: In uniform fields, $d$ is the distance between plates. Do not use the inverse-square law ($1/r^2$) which only applies to point charges (radial fields).
• ❌ Unit Conversion Errors: Forgetting to convert plate separation from $\text{mm}$ or $\text{cm}$ to $\text{m}$. This is the most common cause of power-of-ten errors in 9702 exams.
• ❌ Direction of Electron Motion: Forgetting that electrons are negatively charged. They accelerate opposite to the direction of the electric field lines (toward the positive plate).
• ❌ Ignoring the "Uniform" Requirement: Assuming a field is uniform when it is actually radial (from a point charge). Always check if the question specifies "parallel plates."
• ❌ Vector Signs: In $E = \Delta V / d$, $E$ is the magnitude. When considering motion, ensure the sign of the acceleration matches the direction of the force relative to your chosen coordinate system.
• ❌ Misinterpreting $V$: Using the potential of one plate instead of the potential difference ($\Delta V$) between the two plates. If one plate is $+200\text{ V}$ and the other is $-200\text{ V}$, $\Delta V = 400\text{ V}$, not $200\text{ V}$.

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Exam Tips

1. Describing Uniformity: If asked to describe a uniform field, you must mention two things: the field lines are parallel and they are equally spaced.
2. The "Show That" Question: If a question asks you to "show that" the speed of a particle is a certain value, always start by equating work done to kinetic energy: $q\Delta V = \frac{1}{2}mv^2$.
3. Gravity vs. Electric Force: For subatomic particles (electrons, protons), the gravitational force ($mg$) is almost always negligible compared to the electric force ($qE$). You should only consider gravity if the particle is relatively massive (like an oil drop or a large dust grain) or if the question explicitly mentions "weight."
4. Significant Figures: Always provide your final answer to the same number of significant figures as the least precise data value given in the question (usually 2 or 3 s.f.).
5. Path Visualization: When drawing the path of a particle entering a field perpendicularly, ensure the path is a smooth curve (parabola) while inside the plates and a straight line the moment it exits the plates (where the force becomes zero).
6. Equipotentials: Remember that equipotential lines in a uniform field are always perpendicular to the electric field lines. If the field lines are vertical, the equipotentials are horizontal. The spacing between equipotential lines of equal potential difference (e.g., every $10\text{ V}$) must be equal.