# Motion of Charged Particles in an E Field(OCR A Level Physics)

Author

Ashika

Expertise

## Dielectric Action in a Parallel Plate Capacitor

• Permittivity is the measure of how easy it is to generate an electric field in a certain material
• The relativity permittivity εr is sometimes known as the dielectric constant
• For a given material, it is defined as:

The ratio of the permittivity of a material to the permittivity of free space

• This can be expressed as:

• Where:
• εr = relative permittivity
• ε = permittivity of a material (F m−1)
• ε0 = permittivity of free space (F m1)

• The relative permittivity has no units because it is a ratio of two values with the same unit

• When the polar molecules in a dielectric align with the applied electric field from the plates, they each produce their own electric field
• This electric field opposes the electric field from the plates

The electric field of the polar molecules opposes that of the electric field produced by the parallel plates

• The larger the opposing electric field from the polar molecules in the dielectric, the larger the permittivity
• In other words, the permittivity is how well the polar molecules in a dielectric align with an applied electric field

• The opposing electric field reduces the overall electric field, which decreases the potential difference between the plates
• Therefore, the capacitance of the plates increases

• The capacitance of a capacitor can also be written in terms of the relative permittivity:

• Where:
• C = capacitance (F)
• A = cross-sectional area of the plates (m2)
• d = separation of the plates (m)
• εr = relative permittivity of the dielectric between the plates
• ε0 = permittivity of free space (F m1)

• Capacitor plates are generally square, therefore, if it has a length of L on all sides then the cross-sectional area will be A = L2

A parallel plate capacitor consists of conductive plates each with area A, a distance d apart and a dielectric ε between them

#### Worked example

Calculate the permittivity of a material that has a relative permittivity of 4.5 × 1011.

Step 1: Write down the relative permittivity equation

Step 2: Rearrange for permittivity of the material ε

ε = εrε0

Step 3: Substitute in the values

ε = (4.5 × 1011) × (8.85 × 1012) = 3.9825 = 4 F m1

#### Worked example

A parallel-plate capacitor has square plates of length L separated by distance d and is filled with a dielectric. A second capacitor has square plates of length 3L separated by distance 3d and has air as its dielectric.

Both capacitors have the same capacitance.

Determine the relative permittivity of the dielectric in the first capacitor.

#### Exam Tip

Remember that A, the cross-sectional area, is only for one of the parallel plates. Don't multiply this by 2 for both the plates for the capacitance equation!

## Motion of Charged Particles in an Electric Field

• A charged particle in an electric field will experience a force on it that will cause it to move
• If a charged particle remains still in a uniform electric field
• It will move parallel to the electric field lines (along or against the field lines depending on its charge)
• If a charged particle in motion travels initially perpendicular through a uniform electric field (e.g. between two charged parallel plates)
• It will experience a constant electric force and travel in a parabolic trajectory

The parabolic path of charged particles in a uniform electric field

• The direction of the parabola will depend on the charge of the particle
• A positive charge will be deflected towards the negative plate
• A negative charge will be deflected towards the positive plate

• The force on the particle is the same at all points and is always in the same direction
• Note: an uncharged particle, such as a neutron experiences no force in an electric field and will therefore travel straight through the plates undeflected

• The amount of deflection depends on the following properties of the particles:
• Mass – the greater the mass, the smaller the deflection and vice versa
• Charge – the greater the magnitude of the charge of the particle, the greater the deflection and vice versa
• Speed – the greater the speed of the particle, the smaller the deflection and vice versa

#### Worked example

A single proton travelling with a constant horizontal velocity enters a uniform electric field between two parallel charged plates. The diagram shows the path taken by the proton.

Draw the path taken by a boron nucleus that enters the electric field at the same point and with the same velocity as the proton.

Atomic number of boron = 5

Mass number of boron = 11

Step 1: Compare the charge of the boron nucleus to the proton

• Boron has 5 protons, meaning it has a charge 5 × greater than the proton
• The force on boron will therefore be 5 × greater than on the proton

Step 2: Compare the mass of the boron nucleus to the proton

• The boron nucleus has a mass of 11 nucleons meaning its mass is 11 × greater than the proton
• The boron nucleus will therefore be less deflected than the proton

Step 3: Draw the trajectory of the boron nucleus

• Since the mass comparison is much greater than the charge comparison, the boron nucleus will be much less deflected than the proton
• The nucleus is positively charged since the neutrons in the nucleus have no charge
• Therefore, the shape of the path will be the same as the proton

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