Last lesson the field was just a picture. Now we put a number on it. The field around a point charge follows an inverse-square law: double your distance and the field drops to a quarter, triple it and the field drops to a ninth. One compact equation, $E = \dfrac{kq}{r^2}$, captures exactly how strongly a charge reaches out into the space around it.
The electric field a small distance from a point charge has some value $E$.
Before reading: if you move to twice that distance, do you think the field becomes one-half, one-quarter, or stays the same? Write your guess and a one-sentence reason.
One equation describes how strongly a point charge reaches into space
The strength of the field tells you how hard the charge pushes; it grows with the source charge and shrinks rapidly as you move away.
Electric field strength $E$ at a distance $r$ from a point charge $q$ is given by the NESA formula-sheet equation:
$$E = \frac{kq}{r^2} \qquad k = 8.99 \times 10^{9}\ \text{N m}^2\,\text{C}^{-2}$$
Two things control the field: the charge $q$ (more charge → stronger field, in direct proportion) and the distance $r$ (field falls off as $1/r^2$). Because $r$ is squared, distance matters far more than it first appears:
The electric field strength around a point charge is $E = \dfrac{kq}{r^2}$, with $k = 8.99 \times 10^{9}\ \text{N m}^2\,\text{C}^{-2}$. $E$ is directly proportional to $q$ and follows an inverse-square law in $r$: doubling $r$ divides $E$ by four.
Pause, copy the highlighted equation and rule into your book before moving on.
Did you get this? True or false: moving from a distance $r$ to a distance $3r$ from a point charge reduces the field strength to one-third of its original value.
Putting real numbers into $E = \dfrac{kq}{r^2}$
We just saw the equation and what each part does. That raises a question: how do we actually compute a field strength from a charge and a distance? This card answers it → substitute carefully into $E = \dfrac{kq}{r^2}$ and watch the powers of ten.
Solving for the field is just careful substitution, the only trap is handling the small charges and the squared distance correctly.
Work in SI units: charge in coulombs (C) and distance in metres (m). The field comes out in N C$^{-1}$.
Find the electric field strength $2.0\ \text{cm}$ from a point charge of $+3.0 \times 10^{-9}\ \text{C}$.
To find a point-charge field, convert $r$ to metres, substitute into $E = \dfrac{kq}{r^2}$, and remember to square $r$. The answer is in N C$^{-1}$ and points away from a positive charge, toward a negative one.
Add the highlighted worked method to your notes before the check below.
Quick check: The field strength at distance $r$ from a point charge is $E$. What is the field strength at distance $2r$?
Pattern A, Calculate then reason
A point charge of $+5.0 \times 10^{-9}\ \text{C}$ sits in air. In your book: (1) calculate the field strength at $r = 0.010\ \text{m}$; (2) calculate the field strength at $r = 0.030\ \text{m}$; (3) state the ratio of your two answers and explain in one sentence why it matches the inverse-square law.
Match the pairs: Match each change to its effect on the field strength $E$ around a point charge.
Dot point: Use the equation E = kq/r² to demonstrate that the electric field strength around a point charge follows the inverse square law
Field-line spacing and the $1/r^2$ law are the same story
We just learned to calculate $E$ at any distance. That raises a question: how does the field-line picture from L03 agree with $E \propto 1/r^2$? This card answers it → because the same fixed number of lines spreads over a larger area as $r$ grows.
The reason the field weakens with distance is geometric: the same lines have to cover a sphere whose area grows as $r^2$, so they thin out as $1/r^2$.
A point charge sends out a fixed number of field lines in every direction. Close to the charge these lines are crowded together, a strong field. Further out, the same lines are spread across the surface of a much larger imaginary sphere. Since the surface area of a sphere is proportional to $r^2$, the line density (and therefore the field strength) drops as $1/r^2$. The picture and the equation tell the same story.
Field-line spacing pictures the inverse-square law: a fixed number of lines spreads over a sphere of area $\propto r^2$, so line density, and field strength, falls as $1/r^2$, matching $E = \dfrac{kq}{r^2}$.
Pause, copy the highlighted link between picture and equation into your book before the check below.
Last check: Decide if the statement is true or false.
Field lines crowd together close to a point charge and spread apart further away because the field strength follows an inverse-square law.
Odd one out: three of these quantities appear in $E = kq/r^2$. Which one does not?
Dot point: Use the equation E = kq/r² to demonstrate that the electric field strength around a point charge follows the inverse square law
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 5(4 marks) 1. A point charge of $+4.0 \times 10^{-9}\ \text{C}$ is in air. Calculate the electric field strength at a distance of $3.0\ \text{cm}$ from the charge, and state how the field strength would change if the distance were doubled.
1 mark: convert $r$ to $0.030\ \text{m}$ · 1 mark: correct substitution into $E = \dfrac{kq}{r^2}$ · 1 mark: $E \approx 4.0 \times 10^{4}\ \text{N C}^{-1}$ · 1 mark: doubling $r$ reduces $E$ to one-quarter
ApplyBand 5(3 marks) 2. A point charge of $+8.0 \times 10^{-9}\ \text{C}$ produces an electric field of strength $2.0 \times 10^{4}\ \text{N C}^{-1}$ at a certain point. Calculate the distance of that point from the charge. Take $k = 8.99 \times 10^{9}\ \text{N m}^2\ \text{C}^{-2}$.
1 mark: rearranges $E = kq/r^2$ to $r = \sqrt{kq/E}$ · 1 mark: substitutes correctly · 1 mark: $r = 0.060\ \text{m}$ (2 s.f.)
Dot point: Use the equation E = kq/r² to demonstrate that the electric field strength around a point charge follows the inverse square law
AnalyseBand 5(3 marks) 3. Use $E = kq/r^2$ to explain what happens to the field strength when the distance from a point charge is tripled, and explain how a field-line diagram shows the same result.
1 mark: $E \propto 1/r^2$ so tripling $r$ divides $E$ by $9$ · 1 mark: field lines spread over a larger area as $r$ grows · 1 mark: the area of a sphere grows as $r^2$, so line density falls as $1/r^2$
Dot point: Use the equation E = kq/r² to demonstrate that the electric field strength around a point charge follows the inverse square law
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): Converting the distance, $r = 3.0\ \text{cm} = 0.030\ \text{m}$. Substituting into the field equation: $E = \dfrac{kq}{r^2} = \dfrac{(8.99 \times 10^{9})(4.0 \times 10^{-9})}{(0.030)^2} = \dfrac{35.96}{9.0 \times 10^{-4}} \approx 4.0 \times 10^{4}\ \text{N C}^{-1}$, directed radially outward (positive charge). Because the field follows an inverse-square law, doubling the distance to $0.060\ \text{m}$ divides the field by $2^2 = 4$, so the field strength would fall to about $1.0 \times 10^{4}\ \text{N C}^{-1}$, one-quarter of the original value.
Q2 (3 marks): Rearranging the field equation for the distance gives $r = \sqrt{kq/E}$. Substituting, $kq = (8.99 \times 10^{9})(8.0 \times 10^{-9}) = 71.92$, and dividing by the field strength gives $71.92/(2.0 \times 10^{4}) = 3.596 \times 10^{-3}\ \text{m}^2$. Taking the square root, $r = 6.0 \times 10^{-2}\ \text{m}$, that is $6.0\ \text{cm}$. The square root is the step to watch: because the field depends on $r^2$, finding a distance from a field strength always needs it.
Q3 (3 marks): In $E = kq/r^2$ the distance appears squared in the denominator, so $E$ is inversely proportional to $r^2$. Tripling $r$ multiplies $r^2$ by $9$, so the field strength falls to one ninth of its original value. A field-line diagram shows the same thing geometrically. The same fixed number of lines leaves the charge in all directions, and at a distance $r$ they are spread over a sphere whose surface area is $4\pi r^2$. Tripling $r$ multiplies that area by $9$, so the same number of lines is spread over nine times the area and the line density, which represents field strength, falls to one ninth.
You can now calculate the field strength around a point charge with $E = \dfrac{kq}{r^2}$ and predict how it changes with distance. Next lesson uses this field to find the force on a charge placed in it: $F = qE$.
↩ Back to focus-area hubAt the start you predicted what happens to the field when you double the distance. The answer is it drops to one-quarter, not one-half, because $E \propto 1/r^2$ and $2^2 = 4$. If you predicted one-half, revisit Card 1, the squared distance is what makes the field fall off so fast.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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