Home

Physics

Electric fields

Electric field between two parallel plates

Select Lesson

Exam Board

Select an option

Oscillations

Simple harmonic motion

Displacement, velocity and acceleration for simple harmonic motion

Energy within simple harmonic motion

Damping and resonance

Simple harmonic oscillator and pendulum

Gravitational Fields

Gravitational fields

The law of gravitation

Kepler's laws

Gravitational potential and gravitational potential energy

Nuclear physics

Einstein's mass-energy equation

Nuclear fission and nuclear reactors

Nuclear fusion

Radioactivity

Radioactivity: alpha, beta and gamma radiation

Half-life and radioactive decay

Cosmology

Units for astronomical distances

Hubble's law

The evolution of the Universe

Thermal physics

Measuring temperature and temperature scales

Solids, liquids and gases

Internal energy

Specific latent heat and specific heat capacity

Ideal gases

The ideal gas law

Kinetic theory of gases: Root mean square speed

Black-body radiaton and stellar luminosity

Particle physics

The nuclear model of the atom

Magnitudes of the atom

Fundamental particles

Quarks and anti-quarks

Particle accelerators and detectors

Electromagnetism

Magnetic fields and electromagnetism

Magnetic flux density

Charged particles in a magnetic field

Faraday's and Lenz's Laws

Uses of electromagnetic induction

Alternating current and voltage

Capacitance

Capacitors and capacitance

Energy stored in a capacitor

Charging and discharging capacitors

Electric fields

Coulomb's law

Electric field between two parallel plates

Motion of charged particles in an electric field

Electric potential and potential energy

Circular Motion

Angular velocity and the radian

Centripetal acceleration

Centripetal force

Quantum physics

The photon model

The photoelectric effect

The photoelectric effect equation

Wave-particle duality

Energy levels and spectra

Lenses

Power of a lens

Ray diagrams, the thin lens equation and magnification

Waves

Progressive waves

Waves: displacement-time and distance-time graphs

Refraction and Snell's law

Reflection and the critical angle

Intensity of a wave

Diffraction and polarisation

The principle of superposition

Young's double-slit experiment

Stationary waves

Harmonics

Materials

Hooke's law

Elastic and plastic deformation

Stress, strain and Young's modulus

Stress-strain graphs

Fluids

Density and pressure

Fluid flow and viscosity

Terminal velocity

Electrical circuits

Circuits: diagrams and components

Potential difference and electromotive force

Resistance and Ohm's Law

Resistivity

I-V characteristics

Conductors and semiconductors

Combining resistors and Kirchhoff's Laws

Internal resistance

Potential dividers

Current and charge

Current and charges

Conventional current and electron flow

Mean drift velocity

Newton's laws of motion and momentum

Momentum: definition, conservation and calculations

Collisions in two dimensions

Newton's laws of motion

Energy

Forms of energy, conservation and work done

Kinetic energy and gravitational potential energy

Power and efficiency

Motion

Scalar and vector quantities

Resolving vectors

Displacement and velocity

Acceleration and velocity-time graphs

Equations of motion

Acceleration and free fall

Projectile motion

Mass, weight and centre of mass

Resolving forces

Triangle of forces

The principle of moments

Working as a physicist

Physical quantities and units

How to plan an experiment

Hazards, risks and results

Analysing data

Evaluating data

Explainer Video

Tutor: Lex

Summary

Electric field between two parallel plates

In a nutshell

A uniform field is a field in between two parallel plates where the electric field strength is constant. The electric field between the plates of a capacitor are uniform.

Equations

Description	Equation
Electric field strength	$E=\frac Vd$
Capacitance	$C=\frac {\varepsilon_0A}{d}$
Permittivity	$\varepsilon=\varepsilon_r \varepsilon_0$

Constants

name	symbol	value
$permittivity \ of \ free \ space$	$\varepsilon_0$	$8.85 \times 10^{-12} \, F \, m^{-1}$

Variable definitions

Quantity Name	Symbol	Derived Unit	SI BASE Units
$electric \ field \ strength$	$E$	$N\,C^{-1}$	$kg\,m\,s^{-3}\,A^{-1}$
$potential \ difference$	$V$	$V$	$kg\,m^{2}\,s^{-3}\,A^{-1}$
$distance$	$d$	$m$	$m$
$capacitance$	$C$	$F$	$kg^{-1}\,m^{-2}\,s^4\,A^2$
$area$	$A$	$m^2$	$m^2$
$permittivity$	$\varepsilon$	$F\,m^{-1}$	$kg^{-1}\,m^{-3}\,s^4\,A^2$
$relative\ permittivity$	$\varepsilon_r$

Uniform fields

A uniform field is a field where the field strength is constant anywhere in the field. As the field lines represent the field strength, the lines must be uniformly separated and in the same direction.

For this to happen, the surface must be flat, otherwise the field lines wouldn't be parallel.

The plates between a capacitor can be considered to have a uniform electric field.

Physics; Electric fields; KS5 Year 12; Electric field between two parallel plates

The field strength is the same at all points between the two plates and is given by the following equation, where $d$ is the distance between the plates:

$E=\frac Vd$

Note: The units for electric field strength can also be measured in $V\,m^{-1}$ .

Parallel plate capacitors

The capacitance of a capacitor depends on how easy it is to generate an electric field between the two parallel plates. Capacitance is proportional to the surface area of the overlapping plates and inversely proportional to the distance between them.

Mathematically this can be given as:

$C=\frac {\varepsilon_0A}{d}$

If the plates have a dielectric other than air, then the relative permittivity of that material $\varepsilon$ must used instead. If the relative permittivity is given, then the following equation can be used:

$\varepsilon=\varepsilon_r \varepsilon_0 \newline \\[0.1in]$

Example

Two square pieces of conducting metal, with a measured side of $20 \, cm$ , are placed either side of a pane of glass with a thickness of $4.5 \, mm$ . The glass has a permittivity of $3.54 \times 10^{-11} \, F\,m^{-1}$ . Calculate the capacitance of the two plates.

Firstly, write down the known values:

$l=20\, cm=0.2\, m \newline \\[0.1in]d=4.5 \, mm=4.5 \times 10^{-3} \, m \newline \\[0.1in]\varepsilon=3.54\times 10^{-11} \, F\,m^{-1}$

Next, write down the equations needed and rearrange if necessary:

$C=\frac {\varepsilon A}{d}$

Calculate the area of one of the plates:

$A=l^2 \newline \\[0.1in]A=0.2^2 \newline \\[0.1in]A= 0.04 \, m^2$ 

Then, substitute the values into the equation:

$C=\frac {3.54\times 10^{-11}\times 0.04}{4.5\times 10^{-3}} \newline \\[0.1in]C=3.146666...\times 10^{-10}$

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$capacitance, \ C=3.15\times 10^{-10} \, F = 315 \,pF$

The capacitance of the two plates is $\underline{315 \, pF}.$ 

Learn with Basics

Length:

Unit 1

Static electricity

Unit 2

Electric fields and field lines

Jump Ahead

Optional

Unit 3