A LevelPhysicsAQARevision NotesFields & Their ConsequencesElectric FieldsRadial Electric Field

Radial Electric Field (AQA A Level Physics): Revision Note

Exam code: 7408

Written by: Katie M

Reviewed by: Tim

Updated on 2 October 2026

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Understand this topic with deeper explanations and understanding checks.

Electric field of 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 and the electric field strength equation this can be written as

$E = \frac{F}{q} = \frac{Q}{4 π ε_{0} r^{2}}$

Where:
- $Q$ = the point charge producing the radial electric field (C)
- $r$ = distance from the centre of the charge (m)
- $ε_{0}$ = permittivity of free space (F m⁻¹)
Here, $q$ represents the magnitude of a small positive test charge

This equation shows that in a radial electric field, the electric field strength $E$ :
- is not constant
- follows an inverse square law with distance $r$

Graph of electric field strength E against distance r for a positively charged sphere +Q. E is zero inside the sphere and decreases outside as E is proportional to 1 over r squared. A grey sphere and dashed lines identify the regions. — **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
- If the charge is negative, the E field strength is negative and points towards the centre of the charge
- If the charge is positive, the E field strength is positive and points away from the centre of the charge

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

Two charges on a line with a point X between them show resultant field at X. For +q1 and −q2, electric fields E1 and E2 point right, so Ex = E1 + E2. For two positive charges, E1 points right and E2 left, so Ex = E1 − E2. — **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 $q_{1}$ and $q_{2}$ , with field strengths $E_{1}$ and $E_{2}$ respectively, the magnitude of the resultant field will be:
- the sum of the fields, $E_{1} + E_{2}$ , if they are both in the same direction
- the difference between the fields, $E_{1} - E_{2}$ , 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

Diagram of two positive charges, plus q1 left of point X and plus q2 below it. At X, E1 points right, E2 upwards and resultant electric field Ex is diagonally upwards right. Ex equals the square root of E1 squared plus E2 squared. — **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 metal sphere of diameter 15 cm is 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.

[2]

Answer:

Step 1: List the known quantities

Electric field strength, $E = 1.5 \times 10^{5} V m^{- 1}$
Radius of sphere, $r = \frac{15}{2} = 7.5 cm = 7.5 \times 10^{- 2} m$
Permittivity of free space, $ε_{0} = 8.85 \times 10^{- 12} F m^{- 1}$

Step 2: Write down the equation for electric field strength

$E = \frac{Q}{4 π ε_{0} r^{2}}$

Step 3: Rearrange for charge $Q$

Recall that, for an external observer, a charged sphere can be considered as a point charge
The field strength at the surface of the sphere is equal to the field strength 15 cm away from a point charge with the same total charge as the sphere

$Q = 4 π ε_{0} E r^{2}$

Step 4: Substitute in the values and calculate:

$Q = (4 π \times 8.85 \times 10^{- 12}) \times (1.5 \times 10^{5}) \times {(7.5 \times 10^{- 2})}^{2}$ [1 mark]

$Q = 9.38 \times 10^{- 8} C$

The particle has a charge of 9.4 × 10⁻⁸ C or 94 nC [1 mark]

Examiner Tips and Tricks

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.

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