Electric fields and Coulomb's lawEdexcel International A Level Physics: Revision notes
Section 1
Electric fields and field strength
An electric field is a region where a charged particle experiences a force. The electric field strength E at a point is the force per unit positive charge placed at that point:
E = F/Q
E is a vector, measured in N C⁻¹ (equivalently V m⁻¹). Its direction is the direction of the force on a positive test charge, so it points away from positive charges and towards negative charges. The force on a charge Q in a field E is F = EQ, and it is reversed for a negative charge.
The direction of E is defined by a positive test charge. A negative charge in the field is pushed the opposite way.
Section 2
Coulomb's law
The force between two point charges Q1 and Q2 a distance r apart in a vacuum (or air) is
F = Q1Q2/(4πε0r²)
where ε0 is the permittivity of free space, 8.85 × 10⁻¹² F m⁻¹. The force is attractive for unlike charges and repulsive for like charges, and the two forces on the charges are equal and opposite (Newton's third law), whatever the sizes of the charges.
The force obeys an inverse square law: doubling the separation reduces F to one quarter.
Convert cm, mm and nC to m and C before substituting, and use the data booklet value for ε0.
Section 3
The field of a point charge
Combining E = F/Q with Coulomb's law (using a test charge q) gives the field of a point charge Q:
E = Q/(4πε0r²)
The field is radial and also obeys an inverse square law. It points away from a positive Q and towards a negative Q. A charged sphere has the same field outside it as a point charge of the same charge at its centre.
Where several charges act, find the field from each and add them as vectors (the principle of superposition). Fields in opposite directions subtract.
Do not add field strengths from two charges without checking directions. Between two like charges the fields oppose.
Section 4
Worked example
Charges +4.0 nC and +9.0 nC are 0.30 m apart. Find where the field is zero.
Between like charges the fields oppose, so they can cancel. Let the point be x from the 4.0 nC charge:
4.0/x² = 9.0/(0.30 − x)², so (0.30 − x)/x = 3/2 and x = 0.12 m.
At this point each field is Q/(4πε0r²) = 2.5 × 10³ N C⁻¹, in opposite directions. The zero point is always nearer the smaller charge.
Must know
- E = F/Q, in N C⁻¹; direction of force on a positive charge
- F = Q1Q2/(4πε0r²): attractive for unlike, repulsive for like
- E = Q/(4πε0r²) for a point charge; radial, inverse square
- Fields from several charges add as vectors
- The force on a charge Q in a field is F = EQ
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Electric fields and Coulomb's law
- Two very small charged spheres, A and B, carry charges of +2.0 nC and +5.0 nC respectively. They are held 4.0 cm apart in air, so they may be treated as point charges in a vacuum.Calculate the magnitude of the force between the spheres when they are 4.0 cm apart.2 marks
- In a simple model of a hydrogen atom, an electron moves in a circular orbit around a single proton. The separation of the two particles is 5.3 × 10⁻¹¹ m. The magnitude of the charge on each particle is 1.60 × 10⁻¹⁹ C.Calculate the force on the electron due to the proton and state its direction.2 marks
- A small metal sphere carries a charge of +6.0 nC. Point P is 0.30 m from the centre of the sphere, in air.Calculate the electric field strength at P and state its direction.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).