Uniform electric field (6.2.3)
On this page
When two parallel conducting plates have a potential difference applied across them, a uniform electric field is created in the space between them, pointing from the positive plate to the negative plate.
It is important to note that the electric field strength between two oppositely charged parallel plates is related to the potential difference, applied across them and their separation,
A uniform electric field is represented by parallel, evenly spaced electric field lines. The image below shows a positive test charge placed between two oppositely charged plates:
The positive point charge (blue circle) in the image experiences a constant force and gains kinetic energy as it travels from the positive plate to the negative plate along the electric field lines.
The equation for the electric field strength between two oppositely charged parallel plates can be expressed in terms of the potential difference, between the two plates.
A charge experiences a force when moving between two oppositely charged parallel plates. Work is done on the charge and is equal to the force and the distance the charge moves which is the plate separation.
It is important to note that since the definition of potential difference is the work done per unit charge, we can write the electric strength between two oppositely charged parallel plates as:
The units for the electric field strength are The above equation shows the units for electric field strength are also
This equation is only applicable to oppositely charged parallel plates.
When drawing the electric field lines between two oppositely charged conducting plates, ensure that the field lines:
Question walkthrough
Deriving Electric Field Strength from Potential Difference
Derive an expression for the electric field strength between charged parallel plates in terms of potential difference, combining the definitions of electric field strength, work done, and potential difference.
Question walkthrough
Finding Potential Difference from Force on a Proton
Calculate the potential difference across charged parallel plates given the electrostatic force on a proton and the plate separation, using the relationships between electric field strength, force, and potential difference.
The diagram below shows a parallel plate capacitor, where the space between the two identical plates is a vacuum:
A battery connected to the capacitor creates a potential difference across the two plates, which forces electrons towards one plate and away from the other. The difference in the number of charge carriers between the plates results in two oppositely charged plates, creating an electric field between them.
The strength of the field between the two capacitor plates is dependent on the amount of charge stored on the plates: the more charge, the greater the electric field strength. The electric field strength is also dependent on the distance between the two plates: the shorter the distance, the greater the electric field strength.
It is important to note that capacitance is the amount of charge a component can store at a given potential difference defined by:
Where, in the context of a parallel plate capacitor:
- is the magnitude of charge stored on each plate, and
- is the potential difference between the plates.
The capacitance of a parallel plate capacitor in a vacuum is directly proportional to the area of overlap, since more charge can be stored in a larger area.
The capacitance of a parallel plate capacitor in a vacuum is also inversely proportional to the plate separation, since a smaller plate separation leads to a larger electric field between the plates, meaning a smaller voltage is required to store a given charge.
Therefore, the equation for the capacitance of a parallel plate capacitor is given by:
Where:
- is the capacitance of the capacitor,
- is the area of overlap,
- is the plate separation distance, and
- the proportionality constant is the permittivity of free space, .
When an insulator, also known as a dielectric, is inserted into the space between the plates of a capacitor, the equation for capacitance now incorporates the permittivity of the insulator, .
Where:
- is the relative permittivity,
- is the permittivity of free space.
Therefore, the capacitance of a parallel plate capacitor with a dielectric becomes:
Where:
- is the capacitance of the capacitor,
- is the area of overlap,
- is the plate separation distance,
Different dielectric materials have differing values for the relative permittivity, . For example, a vacuum has a relative permittivity of 1 by definition, and air has a value of 1.0006. The relative permittivity is a dimensionless quantity.
The table below gives the values of the relative permittivity for some common dielectric materials:
The permittivity of a material quantifies its response to an electric field. For the same applied voltage, a material with a higher permittivity has a stronger electric field. Therefore, a smaller voltage is required to store a given charge and this leads to a larger capacitance.
Question walkthrough
Calculating Capacitance of a Dielectric-Filled Capacitor
Calculate the capacitance of a parallel-plate capacitor filled with a perspex dielectric, given the plate dimensions, separation, and relative permittivity.
Electric fields can be used to accelerate charged particles. For example, linear accelerators utilise electric fields to accelerate protons to speeds approaching the speed of light.
The image below shows two oppositely charged plates with a proton placed between them:
The proton, being positively charged, moves downwards following the electric field lines. As the force is constant, the proton has constant acceleration.
When a charged particle travels into a uniform electric field at a right angle, the particle follows a parabolic trajectory.
The image below shows the motion of a positively charged particle entering a uniform electric field at a right angle:
It is important to note that the particle follows a curved trajectory whilst inside the field and continues to follow a straight path once it has left the field.
There is no horizontal acceleration since no horizontal force is being applied, thus the horizontal component of the velocity remains constant throughout the motion.
There is a vertical acceleration of the particle due to the electric field applying a vertically downward force on the particle. The particle initially has zero vertical velocity but upon exiting the field has acquired a vertical component of velocity .
The increase in the vertical velocity of a particle moving through a uniform, vertical electric field with horizontal velocity, can be calculated using Newton’s Second Law of Motion.
The time spent within the field is equal to the length of the electric field region, divided by the particle’s horizontal velocity .
Therefore, the equation for the vertical velocity component of a charged particle moving through a uniform, vertical electric field :
Where:
- is the electric field strength,
- is the partcle’s charge,
- is the mass of the particle, and
- is the length of the electric field region.
If the particle has an initial vertical velocity component equal to then the final vertical velocity is:
Question walkthrough
Deriving Vertical Velocity in an Electric Field
Derive an expression for the vertical velocity gained by a charged particle in a uniform electric field, combining Newton's second law, the electric field strength equation, and the SUVAT equations of motion.
Question walkthrough
Proton Deflection in a Uniform Electric Field
Calculate the vertical velocity component gained by a proton deflected in a uniform electric field between charged parallel plates, using the electric field strength, plate separation, and initial horizontal velocity.