Uniform electric field

What you'll learn

What a uniform electric field is and how to use $E = V / d$ .
How a parallel plate capacitor depends on plate area, separation and permittivity.
How to calculate the force, acceleration and motion of charged particles in a uniform electric field.
How to choose between a force method and an energy method in exam questions.

Prerequisites: charge, potential difference and field strength

A charge is a property of a particle or object that causes it to experience an electric force. Charge is measured in coulombs (C). The symbol for charge is usually $Q$ .

A potential difference is the energy transferred per unit charge between two points. It is measured in volts (V). If a charge moves through a potential difference, the energy transferred is:

W = Q V

where $W$ is work done or energy transferred in joules (J).

Definition

Electric field strength

The electric field strength, $E$ , at a point is the force per unit positive charge placed at that point: $E = F / Q$ . Its unit is newtons per coulomb, $C−1\text{N C}^{-1}$ .

This means a charge in an electric field experiences a force:

F = QE

For a positive charge, the force is in the direction of the electric field. For a negative charge, the force is in the opposite direction.

What makes a field uniform?

A uniform electric field has the same electric field strength and direction at every point in the region being considered.

The clearest A-Level example is the central region between two large, parallel conducting plates connected to a power supply. Field lines are straight, parallel and equally spaced.

Uniform electric field between parallel plates

Definition

Uniform electric field

A uniform electric field is an electric field in which $E$ is constant in magnitude and direction throughout the region.

Between parallel plates, the field points from the positive plate to the negative plate. Ignoring edge effects, its magnitude is:

\frac{V}{d}

where $V$ is the potential difference between the plates in volts (V), and $d$ is the plate separation in metres (m).

Key Idea

Potential gradient

In a uniform field, the electric field strength is the potential difference per metre: $E = V / d$ . The unit $m−1\text{V m}^{-1}$ is therefore equivalent to $C−1\text{N C}^{-1}$ .

You can also understand this from work done. Moving a charge $Q$ across the plates transfers energy $Q V$ . The electric force is $QE$ , and over distance $d$ the work done is $QE d$ . Equating these gives $Q V = QE d$ , so $E = V / d$ .

Example

Finding the field strength between plates

Two parallel plates are separated by 6.0 mm and have a potential difference of 480 V across them. A small positive charge of 2.0 nC is placed between the plates. Find the electric field strength and the force on the charge.

Convert the separation into metres: $\times 10^{-3}\ \text{m}$ .
Use the uniform-field equation:
$\frac{V}{d} = \frac{480\ \text{V}}{6.0 \times 10^{-3}\ \text{m}} = 8.0 \times 10^{4}\ \text{V m}^{-1}$
Use $F = QE$ , with $\times 10^{-9}\ \text{C}$ :
$\times 10^{-9}\ \text{C})(8.0 \times 10^{4}\ \text{N C}^{-1}) = 1.6 \times 10^{-4}\ \text{N}$
Since the charge is positive, the force is in the direction of the field: from the positive plate towards the negative plate.

Common Mistake

Forgetting to convert the gap

The distance $d$ in $E = V / d$ must be in metres. Millimetres and centimetres left unconverted are a very common source of answers that are too large or too small by factors of 1000 or 100.

Common Mistake

Fringing fields

The equation $E = V / d$ assumes the field is uniform, which is a good model in the central region between large parallel plates. Near the edges, field lines curve outwards, so the field is no longer perfectly uniform.

Parallel plate capacitors

A capacitor is a component that stores charge and energy. In its simplest form, it consists of two conducting plates separated by an insulator.

When a potential difference is applied, one plate gains charge $+ Q$ and the other gains charge $- Q$ . The capacitance tells you how much charge is stored per volt:

\frac{Q}{V}

Capacitance is measured in farads (F).

Definition

Parallel plate capacitor

A parallel plate capacitor is made from two parallel conducting plates separated by an insulating material. For large plates with small separation, the field between them is approximately uniform.

For a parallel plate capacitor with vacuum, or approximately air, between the plates:

\frac{\epsilon_0 A}{d}

where $A$ is the overlapping plate area in square metres, $d$ is the separation in metres, and $ϵ0\epsilon_0$ is the permittivity of free space:

m−1\epsilon_0 = 8.85 \times 10^{-12}\ \text{F m}^{-1}

Permittivity and dielectrics

A dielectric is an insulating material placed between the plates of a capacitor. Examples include plastic, oil, glass or ceramic.

Definition

Permittivity

The permittivity, $ϵ\epsilon$ , of a material describes how the material affects the electric field and capacitance. It has unit farads per metre, $m−1\text{F m}^{-1}$ .

For a material between the plates:

\frac{\epsilon A}{d}

The permittivity of the material is related to its relative permittivity, $ϵr\epsilon_r$ , by:

ϵ=ϵrϵ0\epsilon = \epsilon_r \epsilon_0

Relative permittivity has no unit. You are expected to use this relationship, but you are not expected to explain why $ϵr≥1\epsilon_r \ge 1$ .

Key Idea

How to change capacitance

For a parallel plate capacitor, capacitance increases if the plate area $A$ increases, the separation $d$ decreases, or a material with larger permittivity $ϵ\epsilon$ is inserted.

Example

Calculating capacitance with a dielectric

A parallel plate capacitor has plate area 0.020 m² and plate separation 0.50 mm. A dielectric of relative permittivity 2.3 fills the gap. Calculate the capacitance.

Convert the separation into metres and find the material permittivity:
$0.50\ \text{mm} = 5.0 \times 10^{-4}\ \text{m}$ $m−1\epsilon = \epsilon_r \epsilon_0 = (2.3)(8.85 \times 10^{-12}\ \text{F m}^{-1}) = 2.04 \times 10^{-11}\ \text{F m}^{-1}$
Substitute into $\epsilon A/d$ :
$\frac{(2.04 \times 10^{-11}\ \text{F m}^{-1})(0.020\ \text{m}^{2})}{5.0 \times 10^{-4}\ \text{m}}$
Calculate and quote to a sensible number of significant figures:
$\times 10^{-10}\ \text{F}$
This is 0.81 nF.

Tip

Quick proportional checks

For $\epsilon A/d$ , doubling $A$ doubles $C$ , doubling $ϵ\epsilon$ doubles $C$ , but doubling $d$ halves $C$ .

Motion of charged particles in a uniform electric field

A charged particle in a uniform electric field experiences a constant force, provided its charge remains constant and the field is uniform.

Using:

F = QE

and Newton’s second law:

F = ma

the acceleration is:

\frac{QE}{m}

where $m$ is the mass of the particle in kilograms.

For a negative particle, such as an electron, $Q$ is negative, so the acceleration is opposite to the electric field direction. In many calculations, you use the magnitude of the charge to find the size of the acceleration, then state the direction separately.

Motion parallel to the field: energy method

If a charged particle starts from rest and is accelerated through a potential difference, an energy method is often quickest.

If electrical energy becomes kinetic energy:

\frac{1}{2}mv^2

using the magnitude of the charge for energy gained.

Example

Speed after acceleration through a potential difference

An electron is accelerated from rest through a potential difference of 1.5 kV. Calculate its speed. Use $\times 10^{-19}\ \text{C}$ and $kgm_e = 9.11 \times 10^{-31}\ \text{kg}$ .

Convert the potential difference: $\times 10^{3}\ \text{V}$ .
Use energy transfer to kinetic energy:
$\frac{1}{2}m_ev^2$
Rearranging gives:
$\sqrt{\frac{2eV}{m_e}}$
Substitute values:
$\sqrt{\frac{2(1.60 \times 10^{-19}\ \text{C})(1.5 \times 10^{3}\ \text{V})}{9.11 \times 10^{-31}\ \text{kg}}}$
Calculate:
$\times 10^{7}\ \text{m s}^{-1}$

Common Mistake

Confusing energy direction with force direction

For an electron, the force is opposite to the electric field. But when calculating energy gained from an accelerating potential difference, use the magnitude of the charge and decide the direction from the force separately.

Motion perpendicular to the field: projectile-style method

If a charged particle enters a uniform electric field with velocity perpendicular to the field, treat the motion like projectile motion:

horizontal motion: constant velocity, if there is no horizontal electric force;
vertical motion: constant acceleration due to the electric force;
the path is a parabola while the particle is inside the uniform field.

Charged particle deflection in a uniform electric field

Example

Deflection between charged plates

An alpha particle enters horizontally midway between two plates. The plates are 20 mm apart with a potential difference of 200 V. The field region is 30 mm long. The alpha particle has charge $+3.20 \times 10^{-19}\ \text{C}$ , mass $6.64 \times 10^{-27}\ \text{kg}$ , and initial horizontal speed $4.0 \times 10^{5}\ \text{m s}^{-1}$ . Find its vertical deflection as it leaves the plates.

Calculate the electric field strength:
$\frac{V}{d} = \frac{200\ \text{V}}{0.020\ \text{m}} = 1.0 \times 10^{4}\ \text{V m}^{-1}$
Calculate the vertical acceleration:
$\frac{QE}{m} = \frac{(3.20 \times 10^{-19}\ \text{C})(1.0 \times 10^{4}\ \text{N C}^{-1})}{6.64 \times 10^{-27}\ \text{kg}}$ $\times 10^{11}\ \text{m s}^{-2}$
Calculate the time spent between the plates using horizontal motion:
$\frac{L}{u} = \frac{0.030\ \text{m}}{4.0 \times 10^{5}\ \text{m s}^{-1}} = 7.5 \times 10^{-8}\ \text{s}$
Use vertical constant-acceleration motion with initial vertical velocity zero:
$\frac{1}{2}at^2$ $\frac{1}{2}(4.82 \times 10^{11}\ \text{m s}^{-2})(7.5 \times 10^{-8}\ \text{s})^2$
Calculate the deflection:
$\times 10^{-3}\ \text{m}$
The alpha particle is deflected 1.4 mm in the direction of the electric field.

Tip

Choose the method from the motion

If the particle moves along the field, energy is often fastest: $\frac{1}{2}mv^2$ . If it enters across the field, use force and acceleration: $F = QE$ , $F = ma$ , then SUVAT.

Exam technique

In the exam

Always convert distances into metres before using $E = V / d$ or $\epsilon A/d$ .
Decide whether the question is about a field, a capacitor, or particle motion before choosing the equation.
For charged-particle motion, separate direction from magnitude: calculate the size using $F = QE$ or $Q V$ , then use the sign of the charge to state the direction.

Self review

Check yourself

Why are the field lines between ideal parallel plates drawn straight, parallel and equally spaced?
What happens to the capacitance of a parallel plate capacitor if the plate separation is halved?
A negative particle enters a downward electric field moving horizontally. Which way does it accelerate?

What you'll learn

What a uniform electric field is and how to use $E = V / d$ .
How a parallel plate capacitor depends on plate area, separation and permittivity.
How to calculate the force, acceleration and motion of charged particles in a uniform electric field.
How to choose between a force method and an energy method in exam questions.

Prerequisites: charge, potential difference and field strength

A charge is a property of a particle or object that causes it to experience an electric force. Charge is measured in coulombs (C). The symbol for charge is usually $Q$ .

A potential difference is the energy transferred per unit charge between two points. It is measured in volts (V). If a charge moves through a potential difference, the energy transferred is:

W = Q V

where $W$ is work done or energy transferred in joules (J).

Definition

Electric field strength

The electric field strength, $E$ , at a point is the force per unit positive charge placed at that point: $E = F / Q$ . Its unit is newtons per coulomb, $C−1\text{N C}^{-1}$ .

This means a charge in an electric field experiences a force:

F = QE

For a positive charge, the force is in the direction of the electric field. For a negative charge, the force is in the opposite direction.

What makes a field uniform?

A uniform electric field has the same electric field strength and direction at every point in the region being considered.

The clearest A-Level example is the central region between two large, parallel conducting plates connected to a power supply. Field lines are straight, parallel and equally spaced.

Uniform electric field between parallel plates

Definition

Uniform electric field

A uniform electric field is an electric field in which $E$ is constant in magnitude and direction throughout the region.

Between parallel plates, the field points from the positive plate to the negative plate. Ignoring edge effects, its magnitude is:

\frac{V}{d}

where $V$ is the potential difference between the plates in volts (V), and $d$ is the plate separation in metres (m).

Key Idea

Potential gradient

In a uniform field, the electric field strength is the potential difference per metre: $E = V / d$ . The unit $m−1\text{V m}^{-1}$ is therefore equivalent to $C−1\text{N C}^{-1}$ .

Example

Finding the field strength between plates

Convert the separation into metres: $\times 10^{-3}\ \text{m}$ .
Use the uniform-field equation:
$\frac{V}{d} = \frac{480\ \text{V}}{6.0 \times 10^{-3}\ \text{m}} = 8.0 \times 10^{4}\ \text{V m}^{-1}$
Use $F = QE$ , with $\times 10^{-9}\ \text{C}$ :
$\times 10^{-9}\ \text{C})(8.0 \times 10^{4}\ \text{N C}^{-1}) = 1.6 \times 10^{-4}\ \text{N}$
Since the charge is positive, the force is in the direction of the field: from the positive plate towards the negative plate.

Common Mistake

Forgetting to convert the gap

The distance $d$ in $E = V / d$ must be in metres. Millimetres and centimetres left unconverted are a very common source of answers that are too large or too small by factors of 1000 or 100.

Common Mistake

Fringing fields

Parallel plate capacitors

A capacitor is a component that stores charge and energy. In its simplest form, it consists of two conducting plates separated by an insulator.

When a potential difference is applied, one plate gains charge $+ Q$ and the other gains charge $- Q$ . The capacitance tells you how much charge is stored per volt:

\frac{Q}{V}

Capacitance is measured in farads (F).

Definition

Parallel plate capacitor

For a parallel plate capacitor with vacuum, or approximately air, between the plates:

\frac{\epsilon_0 A}{d}

where $A$ is the overlapping plate area in square metres, $d$ is the separation in metres, and $ϵ0\epsilon_0$ is the permittivity of free space:

m−1\epsilon_0 = 8.85 \times 10^{-12}\ \text{F m}^{-1}

Permittivity and dielectrics

A dielectric is an insulating material placed between the plates of a capacitor. Examples include plastic, oil, glass or ceramic.

Definition

Permittivity

The permittivity, $ϵ\epsilon$ , of a material describes how the material affects the electric field and capacitance. It has unit farads per metre, $m−1\text{F m}^{-1}$ .

For a material between the plates:

\frac{\epsilon A}{d}

The permittivity of the material is related to its relative permittivity, $ϵr\epsilon_r$ , by:

ϵ=ϵrϵ0\epsilon = \epsilon_r \epsilon_0

Relative permittivity has no unit. You are expected to use this relationship, but you are not expected to explain why $ϵr≥1\epsilon_r \ge 1$ .

Key Idea

How to change capacitance

For a parallel plate capacitor, capacitance increases if the plate area $A$ increases, the separation $d$ decreases, or a material with larger permittivity $ϵ\epsilon$ is inserted.

Example

Calculating capacitance with a dielectric

A parallel plate capacitor has plate area 0.020 m² and plate separation 0.50 mm. A dielectric of relative permittivity 2.3 fills the gap. Calculate the capacitance.

Convert the separation into metres and find the material permittivity:
$0.50\ \text{mm} = 5.0 \times 10^{-4}\ \text{m}$ $m−1\epsilon = \epsilon_r \epsilon_0 = (2.3)(8.85 \times 10^{-12}\ \text{F m}^{-1}) = 2.04 \times 10^{-11}\ \text{F m}^{-1}$
Substitute into $\epsilon A/d$ :
$\frac{(2.04 \times 10^{-11}\ \text{F m}^{-1})(0.020\ \text{m}^{2})}{5.0 \times 10^{-4}\ \text{m}}$
Calculate and quote to a sensible number of significant figures:
$\times 10^{-10}\ \text{F}$
This is 0.81 nF.

Tip

Quick proportional checks

For $\epsilon A/d$ , doubling $A$ doubles $C$ , doubling $ϵ\epsilon$ doubles $C$ , but doubling $d$ halves $C$ .

Motion of charged particles in a uniform electric field

A charged particle in a uniform electric field experiences a constant force, provided its charge remains constant and the field is uniform.

Using:

F = QE

and Newton’s second law:

F = ma

the acceleration is:

\frac{QE}{m}

where $m$ is the mass of the particle in kilograms.

Motion parallel to the field: energy method

If a charged particle starts from rest and is accelerated through a potential difference, an energy method is often quickest.

If electrical energy becomes kinetic energy:

\frac{1}{2}mv^2

using the magnitude of the charge for energy gained.

Example

Speed after acceleration through a potential difference

An electron is accelerated from rest through a potential difference of 1.5 kV. Calculate its speed. Use $\times 10^{-19}\ \text{C}$ and $kgm_e = 9.11 \times 10^{-31}\ \text{kg}$ .

Convert the potential difference: $\times 10^{3}\ \text{V}$ .
Use energy transfer to kinetic energy:
$\frac{1}{2}m_ev^2$
Rearranging gives:
$\sqrt{\frac{2eV}{m_e}}$
Substitute values:
$\sqrt{\frac{2(1.60 \times 10^{-19}\ \text{C})(1.5 \times 10^{3}\ \text{V})}{9.11 \times 10^{-31}\ \text{kg}}}$
Calculate:
$\times 10^{7}\ \text{m s}^{-1}$

Common Mistake

Confusing energy direction with force direction

Motion perpendicular to the field: projectile-style method

If a charged particle enters a uniform electric field with velocity perpendicular to the field, treat the motion like projectile motion:

horizontal motion: constant velocity, if there is no horizontal electric force;
vertical motion: constant acceleration due to the electric force;
the path is a parabola while the particle is inside the uniform field.

Charged particle deflection in a uniform electric field

Example

Deflection between charged plates

Calculate the electric field strength:
$\frac{V}{d} = \frac{200\ \text{V}}{0.020\ \text{m}} = 1.0 \times 10^{4}\ \text{V m}^{-1}$
Calculate the vertical acceleration:
$\frac{QE}{m} = \frac{(3.20 \times 10^{-19}\ \text{C})(1.0 \times 10^{4}\ \text{N C}^{-1})}{6.64 \times 10^{-27}\ \text{kg}}$ $\times 10^{11}\ \text{m s}^{-2}$
Calculate the time spent between the plates using horizontal motion:
$\frac{L}{u} = \frac{0.030\ \text{m}}{4.0 \times 10^{5}\ \text{m s}^{-1}} = 7.5 \times 10^{-8}\ \text{s}$
Use vertical constant-acceleration motion with initial vertical velocity zero:
$\frac{1}{2}at^2$ $\frac{1}{2}(4.82 \times 10^{11}\ \text{m s}^{-2})(7.5 \times 10^{-8}\ \text{s})^2$
Calculate the deflection:
$\times 10^{-3}\ \text{m}$
The alpha particle is deflected 1.4 mm in the direction of the electric field.

Tip

Choose the method from the motion

If the particle moves along the field, energy is often fastest: $\frac{1}{2}mv^2$ . If it enters across the field, use force and acceleration: $F = QE$ , $F = ma$ , then SUVAT.

Exam technique

In the exam

Always convert distances into metres before using $E = V / d$ or $\epsilon A/d$ .
Decide whether the question is about a field, a capacitor, or particle motion before choosing the equation.
For charged-particle motion, separate direction from magnitude: calculate the size using $F = QE$ or $Q V$ , then use the sign of the charge to state the direction.

Self review

Check yourself

Why are the field lines between ideal parallel plates drawn straight, parallel and equally spaced?
What happens to the capacitance of a parallel plate capacitor if the plate separation is halved?
A negative particle enters a downward electric field moving horizontally. Which way does it accelerate?

What you'll learn

Prerequisites: charge, potential difference and field strength

What makes a field uniform?

Parallel plate capacitors

Permittivity and dielectrics

Motion of charged particles in a uniform electric field

Motion parallel to the field: energy method

Motion perpendicular to the field: projectile-style method

Review Uniform electric field by teaching Genie

Remember key concepts with flashcards

Uniform electric field

What you'll learn

Prerequisites: charge, potential difference and field strength

What makes a field uniform?

Parallel plate capacitors

Permittivity and dielectrics

Motion of charged particles in a uniform electric field

Motion parallel to the field: energy method

Motion perpendicular to the field: projectile-style method

Review Uniform electric field by teaching Genie

Remember key concepts with flashcards

Uniform electric field

What you'll learn

Prerequisites: charge, potential difference and field strength

What makes a field uniform?

Parallel plate capacitors

Permittivity and dielectrics

Motion of charged particles in a uniform electric field

Motion parallel to the field: energy method

Motion perpendicular to the field: projectile-style method

Review Uniform electric field by teaching Genie

Remember key concepts with flashcards

Uniform electric field Revision Guide

Uniform electric field

What you'll learn

Prerequisites: charge, potential difference and field strength

What makes a field uniform?

Parallel plate capacitors

Permittivity and dielectrics

Motion of charged particles in a uniform electric field

Motion parallel to the field: energy method

Motion perpendicular to the field: projectile-style method

Review Uniform electric field by teaching Genie

Remember key concepts with flashcards

Uniform electric field Revision Guide