Key points
- Coulomb's law: F = Qq/4πε₀r², an inverse square law, attractive for unlike charges and repulsive for like.
- Field strength E = F/Q is a vector, in N C⁻¹ or V m⁻¹.
- Point charge: E = Q/4πε₀r². Uniform field between plates: E = V/d.
- Potential V = Q/4πε₀r, and field strength is minus the gradient of the V-r graph.
- A charged particle entering a uniform field at right angles follows a parabola.
- Gravity is always attractive and far weaker. Electric forces can attract or repel.
Coulomb's law and electric field strength in one page
Coulomb's law and electric field strength describe how charges push and pull on each other. Coulomb's law gives the force between two point charges: F = Qq/4πε₀r². Electric field strength is the force per unit positive charge, E = F/Q. Around a point charge E = Q/4πε₀r², and between parallel plates E = V/d.
Electric forces are about as familiar as it gets: static shocks, a balloon sticking to a wall, the force that holds atoms together. The idea of a field lets you deal with them cleanly. Instead of asking what one charge does to another, you ask what the first charge does to the space around it. Any charge you place in that space then feels a force.
The topic sits in the Electric fields part of module 6. It has a lot in common with gravitational fields, so if you have done one, the other will feel familiar, with some important differences.
Coulomb's law
Coulomb's law says the force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation. It acts along the line joining the charges.
F = Qq / (4πε₀r²)
- F = electrostatic force on each charge (N)
- Q, q = the two charges (C)
- r = separation between the centres of the charges (m)
- ε₀ = permittivity of free space, 8.85 × 10⁻¹² F m⁻¹
1/(4πε₀) = 8.99 × 10⁹ N m² C⁻². The equation works for point charges, and also for uniformly charged spheres if r is measured from the centre.
Three things to take from the equation. First, the force is the same size on both charges (Newton's third law), whatever their sizes. Second, doubling r cuts the force to a quarter, which is the inverse square law. Third, like charges repel and unlike charges attract. In calculations, use the magnitudes of the charges to find the size of the force and decide the direction from the signs.
The air between charges has almost the same permittivity as a vacuum, so the equation works for charges in air. In another material, replace ε₀ with ε = εᵣε₀, and the force falls.
Electric field strength
Electric field strength E at a point is the force per unit positive charge placed at that point. It is a vector, and it points in the direction a small positive test charge would be pushed.
E = F / Q
- E = electric field strength (N C⁻¹, or equivalently V m⁻¹)
- F = force on the charge (N)
- Q = the charge feeling the force (C)
Rearranged, F = EQ. A negative charge feels a force opposite to the field direction.
Around a point charge (or outside a uniformly charged sphere), the field is radial. Put F from Coulomb's law into E = F/q and the test charge cancels:
E = Q / (4πε₀r²)
- Q = the charge creating the field (C)
- r = distance from the centre of the charge (m)
E is proportional to 1/r². Doubling the distance reduces E to a quarter.
Between two parallel charged plates, the field in the middle is uniform: the same strength and direction everywhere (apart from near the edges). It is found from the p.d. between the plates and their separation.
E = V / d
- V = p.d. between the plates (V)
- d = separation of the plates (m)
This shows that N C⁻¹ and V m⁻¹ are the same unit. Use whichever suits the question.
Drawing field lines
Field lines show the direction of the field and, by how close together they are, its strength. The rules are short and examiners check each one.
- Lines show the direction of the force on a positive charge, so they leave positive charges and enter negative ones.
- Lines never cross.
- The closer the lines, the stronger the field.
- They meet a conductor's surface at right angles.
- For a uniform field, draw parallel, equally spaced lines with arrowheads.
Between oppositely charged parallel plates the lines go straight from the positive plate to the negative plate, with equal spacing in the middle. At the edges they bow outwards, which is called edge effects or fringing. You can usually ignore it in calculations.
Electric potential and potential energy
Potential gives you an energy view of the same field. If your teacher says this part is beyond what you need, you can skim it, but it helps with the comparison with gravity and with energy questions.
The electric potential V at a point is the work done per unit positive charge in bringing a small positive test charge from infinity to that point. For a point charge, with V = 0 at infinity:
V = Q / (4πε₀r)
- V = electric potential (V)
- Q = the charge creating the field (C)
- r = distance from the charge (m)
V is positive near a positive charge and negative near a negative charge. It falls as 1/r, more slowly than the field falls as 1/r².
The electric potential energy of a charge q at that point is E_p = qV = Qq/(4πε₀r). To move a charge q between two points with a potential difference ΔV, the work done is ΔW = qΔV.
Field and potential are linked. The field strength is the negative gradient of the potential graph: E = −ΔV/Δr. Equally, the area under an E-r graph gives the change in potential. Between parallel plates, V changes uniformly with distance, so E = V/d is the same idea.
Charged particles in a uniform field
A charge Q in a field E feels a force F = EQ, so it accelerates with a = EQ/m. For a uniform field, that acceleration is constant. This is the electric version of a projectile in a gravitational field, and the maths is the same.
Accelerating from rest. A particle of charge q accelerated through a p.d. V gains kinetic energy equal to the work done by the field: ½mv² = qV. For an electron this is ½mₑv² = eV, which is where the electron-volt comes from (1 eV = 1.60 × 10⁻¹⁹ J).
Entering a field at right angles. Suppose an electron enters the space between horizontal plates with a horizontal speed v, and the field is vertical. Then the motion splits into two independent parts, just as for a thrown ball:
- Along the plates: no force, so constant velocity. The time to pass through plates of length L is t = L/v.
- Across the plates: constant acceleration a = eE/mₑ = eV/(mₑd), starting from zero, so the deflection is y = ½at² = eVL²/(2mₑdv²).
Eliminate t and you get y proportional to x², so the path between the plates is a parabola. After the electron leaves the field it moves in a straight line, because there is no longer a force.
In these questions, the electron's weight is almost always negligible. For an electron the electrical force is of the order of 10⁻¹⁵ N, while its weight is about 9 × 10⁻³⁰ N. You can say this in an answer to justify ignoring gravity.
Electric and gravitational fields compared
The two field types have a lot of mathematical structure in common, and OCR likes questions that ask you to compare them. Here are the main points side by side.
| Feature | Gravitational field | Electric field |
|---|---|---|
| Force law | F = Gm₁m₂/r² | F = Qq/(4πε₀r²) |
| Field strength | g = F/m (N kg⁻¹) | E = F/Q (N C⁻¹) |
| Point source | g = GM/r² | E = Q/(4πε₀r²) |
| Force direction | always attractive | attractive or repulsive |
| Uniform field | g near Earth's surface | E = V/d between plates |
| Source property | mass | charge |
| Relative strength | very weak | very strong |
To see how different the strengths are, compare the electric and gravitational forces between two electrons. Each force is proportional to 1/r², so the ratio does not depend on the separation. F(electric)/F(gravitational) = e²/(4πε₀) ÷ Gmₑ² = 4.2 × 10⁴². The electric force is vastly bigger, but on large scales positive and negative charge cancel, so gravity wins for planets and stars.
Link to gravitational fields and circular motion for more. The mathematics of orbits using an inverse square force applies to charged particles too, though a charge in orbit would radiate energy, so the classical picture of an atom fails.
Worked examples
Force between two charges
Two small spheres carrying charges of +3.0 nC and −5.0 nC are 4.0 cm apart in air. Calculate the force between them and state whether it is attractive or repulsive.
- Convert: 3.0 nC = 3.0 × 10⁻⁹ C, 5.0 nC = 5.0 × 10⁻⁹ C, 4.0 cm = 0.040 m.
- F = Qq/(4πε₀r²) = (8.99 × 10⁹ × 3.0 × 10⁻⁹ × 5.0 × 10⁻⁹) / (0.040)²
- F = 1.349 × 10⁻⁷ / 1.6 × 10⁻³ = 8.4 × 10⁻⁵ N.
- The charges are opposite, so the force is attractive.
Answer 8.4 × 10⁻⁵ N, attractive.
Field strength near a point charge
A small sphere carries a charge of +2.0 μC. (a) Calculate the electric field strength 0.25 m from it. (b) A 4.0 nC charge is placed at that point. Calculate the force on it.
- (a) E = Q/(4πε₀r²) = 8.99 × 10⁹ × 2.0 × 10⁻⁶ / (0.25)²
- E = 1.798 × 10⁴ / 0.0625 = 2.88 × 10⁵ N C⁻¹, directed away from the sphere.
- (b) F = EQ = 2.88 × 10⁵ × 4.0 × 10⁻⁹ = 1.15 × 10⁻³ N.
Answer (a) 2.9 × 10⁵ N C⁻¹ away from the sphere. (b) 1.2 × 10⁻³ N.
An electron deflected between plates
Two parallel horizontal plates 5.0 cm apart and 4.0 cm long have a p.d. of 400 V across them. An electron enters midway between the plates, parallel to them, at 2.0 × 10⁷ m s⁻¹. Calculate (a) the field strength, (b) the acceleration of the electron, (c) the time spent between the plates, and (d) its vertical deflection on leaving the plates.
- (a) E = V/d = 400 / 0.050 = 8.0 × 10³ V m⁻¹.
- (b) F = eE = 1.60 × 10⁻¹⁹ × 8.0 × 10³ = 1.28 × 10⁻¹⁵ N. a = F/m = 1.28 × 10⁻¹⁵ / 9.11 × 10⁻³¹ = 1.41 × 10¹⁵ m s⁻².
- (c) t = L/v = 0.040 / 2.0 × 10⁷ = 2.0 × 10⁻⁹ s.
- (d) y = ½at² = ½ × 1.41 × 10¹⁵ × (2.0 × 10⁻⁹)² = 2.8 × 10⁻³ m.
Answer (a) 8.0 × 10³ V m⁻¹. (b) 1.4 × 10¹⁵ m s⁻². (c) 2.0 × 10⁻⁹ s. (d) 2.8 mm towards the positive plate. This is less than the 25 mm to the plate, so the electron does emerge.
Potential and energy
A point charge of +1.5 μC sits in air. A charge of +2.0 nC is moved from 0.30 m to 0.10 m from it. Calculate the potential at each point and the work done.
- V = Q/(4πε₀r). At 0.30 m: V = 8.99 × 10⁹ × 1.5 × 10⁻⁶ / 0.30 = 4.5 × 10⁴ V.
- At 0.10 m: V = 8.99 × 10⁹ × 1.5 × 10⁻⁶ / 0.10 = 1.35 × 10⁵ V.
- ΔV = 1.35 × 10⁵ − 4.5 × 10⁴ = 9.0 × 10⁴ V.
- W = qΔV = 2.0 × 10⁻⁹ × 9.0 × 10⁴ = 1.8 × 10⁻⁴ J.
Answer 4.5 × 10⁴ V and 1.35 × 10⁵ V. The work done on the charge is 1.8 × 10⁻⁴ J, because the charges repel and you must push it closer.
Common mistakes
- Forgetting to square r. The 1/r² is the whole point of the law. Do the division by r² last and check the power of ten.
- Using centimetres or nanocoulombs unconverted. 4.0 cm = 0.040 m, and 3.0 nC = 3.0 × 10⁻⁹ C.
- Using the radius of the sphere instead of the distance from its centre. For charged spheres, measure r from the centre.
- Treating E as a scalar. When two charges contribute, add the fields as vectors. Fields in opposite directions subtract.
- Mixing up E = F/Q with E = V/d. The first is the definition. The second applies only to a uniform field between plates.
- Saying an electron moves along the field lines. Negative charges move against the field direction.
- Calling the path in a uniform field a circle. A constant force in a fixed direction gives a parabola. A circle needs a force that changes direction, as in a magnetic field.
How OCR A examines this topic
Electric fields are assessed in Module 6, so expect questions in the papers that cover it and in the synoptic paper. Check your own copy of the specification for exact paper coverage. Typical questions include force and field strength calculations, drawing and interpreting field line patterns, charged particles moving between plates, and comparing electric with gravitational fields.
In multi-step questions, marks are given for each stage: converting units, substituting into the right equation, getting the right power of ten, and giving the unit. For deflection questions, a clear statement of the separation into horizontal and vertical motion often earns a mark of its own.
Synoptic links. The motion of charged particles uses the same ideas as projectile motion. The energy gained by a charge accelerated through a p.d. links to kinetic energy and to the photoelectric effect, where electrons are stopped by a potential. Field between plates connects to capacitors in series and parallel.
Practice questions
Two point charges of +2.0 μC are 0.30 m apart in air. Calculate the force between them and state whether it is attractive or repulsive.3 marks
Show the answer
Answer: 0.40 N, repulsive.
F = 8.99 × 10⁹ × (2.0 × 10⁻⁶)² / (0.30)² = 8.99 × 10⁹ × 4.0 × 10⁻¹² / 0.090 = 0.400 N. Like charges repel.
Calculate the electric field strength 0.25 m from a point charge of −8.0 nC, and state its direction.3 marks
Show the answer
Answer: 1.2 × 10³ N C⁻¹, directed towards the charge.
E = 8.99 × 10⁹ × 8.0 × 10⁻⁹ / (0.25)² = 71.9 / 0.0625 = 1.15 × 10³ N C⁻¹. The charge is negative, so the field points towards it.
A p.d. of 1.2 kV is applied across plates 8.0 mm apart. Calculate the field strength and the force on an electron between the plates.3 marks
Show the answer
Answer: E = 1.5 × 10⁵ V m⁻¹ and F = 2.4 × 10⁻¹⁴ N.
E = V/d = 1200 / 8.0 × 10⁻³ = 1.5 × 10⁵ V m⁻¹. F = eE = 1.60 × 10⁻¹⁹ × 1.5 × 10⁵ = 2.4 × 10⁻¹⁴ N.
Calculate the electron acceleration in the field in the previous question, and explain why the weight of the electron can be ignored.3 marks
Show the answer
Answer: a = 2.6 × 10¹⁶ m s⁻². The weight is about 9 × 10⁻³⁰ N, far smaller than the electric force.
a = F/m = 2.4 × 10⁻¹⁴ / 9.11 × 10⁻³¹ = 2.63 × 10¹⁶ m s⁻². Weight = mg = 9.11 × 10⁻³¹ × 9.81 = 8.9 × 10⁻³⁰ N, which is about 10¹⁶ times smaller.
Calculate the distance from a +5.0 μC point charge at which the field strength is 1000 N C⁻¹.3 marks
Show the answer
Answer: 6.7 m.
E = Q/(4πε₀r²) so r² = 8.99 × 10⁹ × 5.0 × 10⁻⁶ / 1000 = 44.95, and r = 6.7 m.
Calculate the electric potential 0.20 m from a +4.0 nC point charge, and the work done in bringing a +1.0 nC charge from far away to that point.3 marks
Show the answer
Answer: V = 180 V and W = 1.8 × 10⁻⁷ J.
V = 8.99 × 10⁹ × 4.0 × 10⁻⁹ / 0.20 = 180 V. W = qV = 1.0 × 10⁻⁹ × 180 = 1.8 × 10⁻⁷ J.
Next steps
Compare what you have learned here with gravitational fields, then look at how the field between plates is used in capacitors in series and parallel. Magnetic fields are next: see magnetic fields and flux density. All the topics are on the module 6 hub. When you are ready, practise this topic with step-by-step marking. It is free to start.
Frequently asked questions
What is Coulomb's law?
Coulomb's law says the force between two point charges is F = Qq/4πε₀r². It is proportional to the product of the charges and inversely proportional to the square of their separation. The force is repulsive for like charges and attractive for unlike charges.
What is the difference between electric field strength and force?
Force depends on the charge placed in the field, while field strength is the force per unit charge, E = F/Q. Field strength tells you about the field itself, whatever charge is placed in it. It is a vector that points in the direction of the force on a positive charge.
Why is the field between parallel plates uniform?
Away from the edges, the field lines are parallel and equally spaced, so the strength and direction are the same everywhere. The field strength is found from E = V/d, the p.d. divided by the plate separation.
Why does a charged particle follow a parabola in a uniform field?
The force on it is constant and in a fixed direction. This gives a constant acceleration at right angles to the initial velocity. Horizontal motion is at constant speed and vertical motion is uniformly accelerated, so y is proportional to x².
How are electric and gravitational fields similar?
Both follow an inverse square law and have a field strength defined as force per unit property (mass or charge). The main differences are that gravity is always attractive, while electric forces can attract or repel, and that gravity is far weaker.
Related guides
Written for the OCR A Level Physics A (H556) specification. Questions are original and written in exam style; they are not taken from OCR papers. Spotted a mistake? Tell us.