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=dV |
Capacitance | C=dε0A |
Permittivity | ε=εrε0 |
Constants
name | symbol | value |
permittivity of free space | | 8.85×10−12Fm−1 |
Variable definitions
Quantity Name | Symbol | Derived Unit | SI BASE Units |
electric field strength | | NC−1 | kgms−3A−1 |
potential difference | | | kgm2s−3A−1 |
distance | | | |
capacitance | | | kg−1m−2s4A2 |
| | | |
permittivity | | | kg−1m−3s4A2 |
relative permittivity | |
|
|
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.
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=dV
Note: The units for electric field strength can also be measured in Vm−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=dε0A
If the plates have a dielectric other than air, then the relative permittivity of that material ε must used instead. If the relative permittivity is given, then the following equation can be used:
ε=εrε0
Example
Two square pieces of conducting metal, with a measured side of 20cm, are placed either side of a pane of glass with a thickness of 4.5mm. The glass has a permittivity of 3.54×10−11Fm−1. Calculate the capacitance of the two plates.
Firstly, write down the known values:
l=20cm=0.2md=4.5mm=4.5×10−3mε=3.54×10−11Fm−1
Next, write down the equations needed and rearrange if necessary:
C=dεA
Calculate the area of one of the plates:
A=l2A=0.22A=0.04m2
Then, substitute the values into the equation:
C=4.5×10−33.54×10−11×0.04C=3.146666...×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=3.15×10−10F=315pF
The capacitance of the two plates is 315pF.