The big idea: An electric field is a region where a charge feels a force.
Electric field strength E measures how strong that field is: the force per unit charge — the force on a tiny test charge, divided by the size of that charge.
It is a vector (it has a direction), and its unit is N C⁻¹ (newtons per coulomb).
[Diagram: phys-field-lines] - Available in full study mode
Which way does the field point?: The field points the way a small positive test charge would be pushed.
So field lines point OUT of a positive charge and IN to a negative charge.
Electric field strength is the force on a test charge divided by the size of that charge:
- electric field strength (N C⁻¹)
- force on the test charge (N)
- size of the small test charge (C)
[Diagram: phys-formula-triangle] - Available in full study mode
The field of a point charge: A single point charge Q makes a field that gets weaker with distance. Combining the data-booklet equations gives:
E = kQ ÷ r²
Double the distance r and the field drops to a quarter (inverse-square). Here k is the Coulomb constant, 8.99 × 10⁹ N m² C⁻².
- electric field strength (N C⁻¹)
- Coulomb constant, 8.99 × 10⁹ N m² C⁻² (given)
- size of the charge making the field (C)
- distance from that charge to the point (m)
Worked example — field of a point charge
A point charge of +2.0 × 10⁻⁶ C sits in a vacuum. Find the electric field strength at a point 0.30 m away. (k = 8.99 × 10⁹ N m² C⁻².)
Solution
- Start with the field of a point charge:
- Put in the numbers (Q = 2.0 × 10⁻⁶, r = 0.30):
- Work it out — keep the unit:
Final answer
E = 2.0 × 10⁵ N C⁻¹, pointing away from the positive charge.
Practice with real exam questions
Answer exam-style questions and get AI feedback that shows you exactly what examiners want to see in a full-marks response.
How this is tested: Field strength and superposition are the core skill in the electric-field questions.
- Paper 1A: find the resultant field at a point between two charges — work out each field with E = kQ ÷ r², then add them as vectors (mind the directions). - Paper 2: locate the zero-field (null) point between two like charges, where the two fields are equal and opposite so they cancel.
Classic trap: adding the two field magnitudes without checking direction. Between two like charges the fields point opposite ways, so you subtract; outside, or between two opposite charges, they point the same way, so you add.
Superposition — add fields as vectors: The total field at a point is the vector sum of the field from each charge.
Work out each one with E = kQ ÷ r², then combine with directions: same direction → add the sizes; opposite directions → subtract them.
IB-style question — (a) field from each charge
Two charges sit on a line 0.40 m apart: a +3.0 × 10⁻⁹ C charge on the left and a +3.0 × 10⁻⁹ C charge on the right. Find the field strength each charge produces at the midpoint, 0.20 m from each. (k = 8.99 × 10⁹ N m² C⁻².)
Solution
- Use the point-charge field for one charge:
- Put in the numbers (Q = 3.0 × 10⁻⁹, r = 0.20):
- Work it out — the same for each charge (same Q, same r):
Final answer
Each charge gives 6.7 × 10² N C⁻¹ at the midpoint.
IB-style question — (b) the resultant field
Using part (a), find the resultant (total) electric field strength at the midpoint between the two equal positive charges.
Solution
- At the midpoint each charge pushes a positive test charge away from itself — so the two fields point opposite ways.
- Opposite directions and equal sizes, so they cancel:
- So the resultant field is:
Final answer
Enet = 0 at the midpoint — the two equal, opposite fields cancel. That midpoint is the zero-field (null) point.