Radial Electric Fields Revision Notes | A Level AQA Physics

1. Why Radial Fields Matter

Every charged particle — a proton, an ion, a charged sphere — creates an electric field around it. That field is not flat or uniform; it fans outward in all directions like light from a bulb. Understanding its shape and strength is the key to predicting how charges interact at any distance.

This lesson focuses on the quantitative description of that field: a single, elegant equation that tells you exactly how strong the field is at any point around a point charge. Once you have this equation, you can analyse satellite-like orbits of charged particles, explain why lightning rods work, and tackle a whole class of AQA exam questions with confidence.

2. The Shape of a Radial Field

Picture a single positive point charge sitting in empty space. Now imagine drawing arrows showing the direction a tiny positive test charge would be pushed at every surrounding point. Those arrows all point directly away from the central charge, spreading outward in three dimensions — this is a radial field.

For a negative point charge, the arrows reverse: they all point inward toward the charge. The geometry is still radial — the field lines are still straight and pass through the central charge — but the direction flips. This is the first key distinction from a uniform field, where all field lines are parallel and equally spaced.

• Radial field: field lines spread outward (or inward) from a central point charge; field strength decreases with distance.
• Uniform field: field lines are parallel and equally spaced; field strength is constant throughout the region.
• In a radial field, the spacing between field lines increases with distance — this directly represents the weakening of the field.

The field line diagram is not decoration — it encodes real physics. Where field lines are close together, the field is strong. Where they spread apart, the field is weak. In a radial field, the lines diverge as you move away from the charge, so the field inevitably weakens with distance. The question is: how does it weaken? That is where the equation comes in.

3. The Equation for Electric Field Strength in a Radial Field

The field strength at a distance $r$ from a point charge $Q$ is given by a relationship derived from Coulomb's law. The force on a small positive test charge $q$ placed at distance $r$ is $F = \frac{1}{4\pi\varepsilon_0}\frac{Qq}{r^2}$. Dividing both sides by $q$ (since $E = F/q$) gives the field strength at that point — independent of the test charge itself.

Notice the $r^2$ in the denominator. This is the inverse-square law at work: double the distance and the field strength drops to one quarter; triple the distance and it drops to one ninth. The field weakens rapidly as you move away from the charge, which matches the spreading-out of field lines you saw in the diagram.

The constant $\frac{1}{4\pi\varepsilon_0}$ appears because the field spreads over the surface area of a sphere ($4\pi r^2$) as it radiates outward — the $4\pi$ is geometric, not arbitrary. You will be given the value of $\varepsilon_0$ in the AQA data booklet, so you do not need to memorise it, but you must know what it represents and how to use it.

4. Worked Calculations

Theory becomes exam-ready through practice. The calculation below walks through a standard AQA-style question step by step. Try each step yourself before revealing the working — this is where the understanding locks in.

Worked example: Calculating E at a distance from a point charge

A point charge of $Q = +3.2 \times 10^{-9}$ C is in a vacuum. Calculate the magnitude of the electric field strength at a distance of $r = 0.040$ m from the charge. Give your answer to 2 significant figures. ($\varepsilon_0 = 8.85 \times 10^{-12}$ F m⁻¹)

1. $E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$ — [M1] Stating the correct equation earns the method mark. Always write it out in full before substituting.
2. $\frac{1}{4\pi\varepsilon_0} = \frac{1}{4 \times \pi \times 8.85 \times 10^{-12}} = \frac{1}{1.113 \times 10^{-10}} \approx 8.99 \times 10^{9}$ N m² C⁻² — You can use this value directly from the AQA data booklet or compute it. Keep it as $8.99 \times 10^9$ for the substitution.
3. $r^2 = (0.040)^2 = 0.040 \times 0.040 = 1.6 \times 10^{-3}$ m² — Expand the square explicitly: $0.040 \times 0.040 = 0.0016 = 1.6 \times 10^{-3}$. Never skip this step — a common error is writing $r^2 = 0.040$ instead of squaring it.
4. $E = 8.99 \times 10^{9} \times \frac{3.2 \times 10^{-9}}{1.6 \times 10^{-3}}$ — [M1] Correct substitution with units. Check that Q is in coulombs and r is in metres before substituting.
5. $\frac{3.2 \times 10^{-9}}{1.6 \times 10^{-3}} = 2.0 \times 10^{-6}$ C m⁻² — $3.2 \div 1.6 = 2.0$ and $10^{-9} \div 10^{-3} = 10^{-9-(-3)} = 10^{-6}$.
6. $E = 8.99 \times 10^{9} \times 2.0 \times 10^{-6} = 17.98 \times 10^{3} \approx 1.8 \times 10^{4}$ N C⁻¹ — [A1] $8.99 \times 2.0 = 17.98$ and $10^9 \times 10^{-6} = 10^3$. Rounded to 2 significant figures: $1.8 \times 10^4$ N C⁻¹.

Answer: $E \approx 1.8 \times 10^4$ N C⁻¹

Notice that the answer is positive — field strength is always quoted as a magnitude. The direction (away from or toward the charge) is determined separately by the sign of $Q$ and described in words or shown on a diagram.

5. How E Varies with Distance: The Inverse-Square Law

The relationship $E \propto \frac{1}{r^2}$ is one of the most important patterns in A-level physics. You have already met it in the context of gravitational fields — the gravitational field strength $g = \frac{GM}{r^2}$ follows exactly the same mathematical form. This is not a coincidence; both fields spread over the surface of a sphere as they radiate outward.

On a graph of $E$ against $r$, the curve is a hyperbola that falls steeply near the charge and flattens out at large distances, never quite reaching zero. On a graph of $E$ against $\frac{1}{r^2}$, the relationship becomes a straight line through the origin — a useful check in data analysis questions.

The steep drop at small $r$ is physically significant: very close to a point charge, the field is enormous. This is why sharp points on conductors concentrate charge and produce very strong local fields — a principle behind lightning conductors and the operation of Van de Graaff generators.

6. Comparing Radial and Uniform Electric Fields

AQA questions frequently ask you to compare these two field types. The table below consolidates the key distinctions — each row maps directly to a potential exam mark.

When an exam question asks you to 'compare', you must address both similarities and differences. Both field types exert a force on a charge placed within them; both can be represented by field lines. The critical difference is that $E$ is position-dependent in a radial field but position-independent in a uniform field.

7. Exam Technique and Common Mistakes

The calculation marks in radial field questions are straightforward to collect if you follow a disciplined method. The marks that students drop are almost always due to unit errors, forgetting to square $r$, or confusing the direction of the field with its magnitude. The guidance below targets those specific traps.

For extended-response questions asking you to 'explain how electric field strength varies with distance from a point charge', a full-mark answer links three things: the equation ($E \propto \frac{1}{r^2}$), the physical reason (field spreads over an ever-larger spherical surface), and the graphical representation (a curve that decreases steeply then flattens, never reaching zero).