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 $\underline{315 \, pF}.$ 