Electric Field Strength

An electric field is a region of space in which an electric charge experiences a force
The electric field strength at a point is defined as:

The force per unit charge experienced by a small positive test charge placed at that point

The electric field strength can be calculated using the equation:

$E = \frac{F}{q}$

Where:
- E = electric field strength (N C⁻¹)
- F = electric force on the charge (N)
- q = magnitude of the charge (C)

Note that the definition specifies that a positive test charge is used
This sets a clear convention for the direction of an electric field, for example, in a field of strength :
- A positive charge $+ q$ experiences a force $E q$ in the direction of the field
- A negative charge $- q$ experiences a force $E q$ in the opposite direction

Hence, electric field strength is a vector quantity and is always directed:
- Away from a positive charge
- Towards a negative charge

Electric Field Strength due to a Point Charge

The strength of an electric field due to a point charge decreases with the square of the distance
- This is an inverse square law, similar to Coulomb's law

Using Coulomb's law, this can be written as

$E = \frac{F}{q} = \frac{k q}{r^{2}}$

Where k = Coulomb constant (N m² C^–2)
A charged sphere acts the same as a point charge, with the same charge as the sphere, at the sphere's centre
- Within the sphere, however, the electric field strength is zero

This means that the electric field of a charged sphere, outside the sphere, is identical to that of a point charge

Graph of field strength against distance for a positive charge

Electric field strength is zero inside a charged sphere and decreases with distance outside the sphere according to an inverse square law

Combining Electric Fields

Both electric force and field strength are vector quantities
Therefore, to find the electric force or field strength at a point due to multiple charges, each field can be combined by vector addition

Vector addition of electric field along the same line

For charges along the same line, the resultant field is the vector addition of the field due to both charges at a particular point

For a point on the same line as two charges q₁ and q₂, with field strengths E₁ and E₂ respectively, the magnitude of the resultant field will be:
- The sum of the fields, E₁ + E₂, if they are both in the same direction
- The difference between the fields, E₁ − E₂, if they are in opposite directions

The direction of the resultant field depends on
- the types of charge (positive or negative)
- the magnitude of the charges

For a point which makes a right-angled triangle with the charges, the resultant field can be determined using Pythagoras theorem

Vector addition of electric field components

4-2-5-vector-addition-of-electric-field-strength-with-pythagoras

For charges which make a right-angle triangle with point X, the resultant field is the vector addition of the field due to both charges using Pythagoras theorem

Worked example

A charged particle experiences a force of 0.3 N at a point where the magnitude of electric field strength is 3.5 × 10⁴ N C⁻¹.

Calculate the magnitude of the charge on the particle.

Answer:

Step 1: Write down the equation for electric field strength

$E = \frac{F}{q}$

Step 2: Rearrange for charge Q

$q = \frac{F}{E}$

Step 3: Substitute in the values and calculate:

$q = \frac{0.3}{3.5 \times 10^{4}} = 8.571 \times 10^{- 6} = 8.6 \times 10^{- 6}$ C (2 s.f.)

The particle has a charge of 8.6 × 10⁻⁶ C or 8.6 μC

Worked example

A metal sphere of diameter 15 cm is uniformly negatively charged. The electric field strength at the surface of the sphere is 1.5 × 10⁵ V m⁻¹.

Determine the total surface charge of the sphere.

Answer:

Step 1: List the known quantities

Electric field strength, E = 1.5 × 10⁵ V m⁻¹
Radius of sphere, r = 15 / 2 = 7.5 cm = 7.5 × 10⁻² m
Coulomb constant, k = 8.99 × 10⁹ N m² C^–2

Step 2: Write down the equation for electric field strength

$E = \frac{k q}{r^{2}}$

It is possible to treat the sphere as a point charge with the same total charge, as it is uniformly charged

Step 3: Rearrange for charge Q

$q = \frac{E r^{2}}{k}$

Step 4: Substitute in the values and calculate:

$q = \frac{(1.5 \times 10^{5}) \times {(7.5 \times 10^{- 2})}^{2}}{8.99 \times 10^{9}} = 9.38 \times 10^{- 8}$ C

The sphere has a charge of 9.4 × 10⁻⁸ C or 94 nC

Examiner Tip

When combining electric fields from multiple charges, remember that the point (e.g. point X in the examples above) represents a positive test charge, so the direction of the electric force or field will correspond to the signs of the charges; the direction of the force or field points away from a positive charge and towards a negative charge.

Electric Field Between Parallel Plates

The magnitude of the electric field strength in a uniform field between two charged parallel plates is defined as:

$E = \frac{V}{d}$

Where:
- E = electric field strength (V m⁻¹)
- V = potential difference between the plates (V)
- d = separation between the plates (m)

Note: both units for electric field strength, V m⁻¹ and N C⁻¹, are equivalent
The equation shows:
- The greater the voltage between the plates, the stronger the field
- The greater the separation between the plates, the weaker the field

This equation cannot be used to find the electric field strength around a point charge
- This is because the field around a point charge is radial

The electric field between two plates is directed:
- From the positive plate (i.e. the one connected to the positive terminal)
- To the negative plate (i.e. the one connected to the negative terminal)

Uniform Electric Field Between two Parallel Plates

Electric field between two plates, downloadable AS & A Level Physics revision notes

The electric field strength between two charged parallel plates is the ratio of the potential difference and separation of the plates

Worked example

Two parallel metal plates separated by 3.5 cm have a potential difference of 7.9 kV between them.

Calculate the electric force acting on a point charge of 2.6 × 10⁻¹⁵ C when placed between the plates.

Answer:

Step 1: List the known quantities

Potential difference between plates, V = 7.9 kV = 7900 V
Distance between plates, d = 3.5 cm = 0.035 m
Charge, q = 2.6 × 10⁻¹⁵ C

Step 2: Equate the equations for electric field strength

E field between parallel plates: $E = \frac{V}{d}$

E field on a point charge: $E = \frac{F}{q}$

$E = \frac{F}{q} = \frac{V}{d}$

Step 3: Rearrange the expression for electric force F

$F = \frac{q V}{d}$

Step 4: Substitute values to calculate the force on the point charge

$F = \frac{(2.6 \times 10^{- 15}) \times 7900}{0.035} = 5.9 \times 10^{- 10}$ N (2 s.f.)

Examiner Tip

Remember the equation for electric field strength with V and d is only valid for parallel plates, and not for point charges

However, when a point charge moves between two parallel plates, the two equations for electric field strength can be equated:

$E = \frac{F}{Q} = \frac{V}{d}$

Top tip: if one of the parallel plates is earthed, it has a voltage of 0 V

Electric Field Strength (HL IB Physics)

Revision Note

Electric Field Strength

Electric Field Strength due to a Point Charge

Graph of field strength against distance for a positive charge

Combining Electric Fields

Vector addition of electric field along the same line

Vector addition of electric field components

Worked example

Worked example

Examiner Tip

Electric Field Between Parallel Plates

Uniform Electric Field Between two Parallel Plates

Worked example

Examiner Tip

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Author: Ann H