E and V Relationships Revision Notes | A Level AQA Physics

1. Why E and V Are Two Sides of the Same Coin

Picture a positive point charge sitting in space. It creates an electric field that pushes other positive charges away — and it also creates a landscape of electric potential around it, like a hill that gets steeper the closer you get. These two quantities, $E$ and $V$, are not independent. They are mathematically linked, and understanding that link is the heart of this topic.

The key insight is this: $E$ tells you how steeply $V$ changes with position. If you know how $V$ varies with $r$, you can find $E$ — and vice versa. The AQA specification asks you to work fluently in both directions, reading graphs and performing calculations. Let's build that fluency from the ground up.

2. The Shapes of E-r and V-r Graphs

Start with the equations you already know for a point charge of magnitude $Q$, where $k = 8.99 \times 10^9$ N m² C⁻² (Coulomb's constant) and $r$ is the distance from the centre of the charge:

These two equations produce very different graph shapes, and examiners test whether you can distinguish them. The $E$-$r$ graph follows a $1/r^2$ curve — it falls steeply at first and then flattens out, always remaining positive for a positive charge. The $V$-$r$ graph follows a $1/r$ curve — it also falls and flattens, but more gradually, and it too remains positive for a positive charge (negative for a negative charge).

Crucially, neither graph ever touches zero at a finite distance — both $E$ and $V$ only reach zero at $r = \infty$. This is the reference point: electric potential is defined as zero at infinity.

3. The Relationship E = ΔV/Δr

Here is where the two quantities connect mathematically. Consider moving a small positive test charge a distance $\Delta r$ outward from a positive point charge. The field does positive work on the test charge as it moves outward (away from the positive source charge), and the electric potential $V$ decreases with increasing $r$. The magnitude of the potential change per unit displacement equals the force per unit charge — that is, the field strength $E$. This gives the fundamental relationship:

For a positive test charge moving outward from a positive source charge, the field does positive work; the physical interpretation is that $E$ is the potential gradient in magnitude form. A region where $V$ changes rapidly with $r$ has a strong field; a region where $V$ barely changes has a weak field. This is why field lines crowd together near a charge — the potential is changing steeply there.

On a $V$-$r$ graph, you find $E$ at a particular point by drawing a tangent to the curve at that point and calculating the gradient of that tangent. The steeper the tangent, the larger $E$. This is a direct graphical application of $E = \Delta V / \Delta r$.

4. Finding ΔV from the Area Under an E-r Graph

The relationship $E = \Delta V / \Delta r$ can be rearranged to give $\Delta V = E \, \Delta r$. For a small strip of width $\Delta r$ under the $E$-$r$ graph, $E \, \Delta r$ is the area of that strip. Adding up all the strips between two distances gives the total change in potential between those distances — the area under the $E$-$r$ graph equals $\Delta V$.

This is the graphical inverse of the previous section. Where $E = \Delta V / \Delta r$ lets you extract $E$ from the gradient of a $V$-$r$ graph, the area rule lets you extract $\Delta V$ from the area under an $E$-$r$ graph. The two operations are mathematical inverses — differentiation and integration — though AQA does not require you to use calculus notation explicitly.

• To find E at a point: draw a tangent to the V-r curve at that point and calculate its gradient (ΔV/Δr).
• To find ΔV between two distances: calculate the area under the E-r graph between those two distances.
• For a curved E-r graph, estimate the area by counting squares or using the trapezium rule.
• Units check: E in V m⁻¹ multiplied by Δr in m gives ΔV in V — always verify units.

5. Worked Calculations

Theory becomes exam-ready through practice. The following two worked examples cover the two core skills: using $E = \Delta V / \Delta r$ to find field strength from a graph gradient, and using the area rule to find $\Delta V$. Work through each step before revealing the next.

Worked example: Finding E from the Gradient of a V-r Graph

A V-r graph for a point charge shows that V = 450 V at r = 0.20 m and V = 180 V at r = 0.50 m. Calculate the average electric field strength E between these two points. [3 marks]

1. $E = \frac{\Delta V}{\Delta r}$ — [M1] Correct formula stated. Note: we use the magnitude of the gradient here.
2. $\Delta V = 450 - 180 = 270 \text{ V}$ — ΔV is the drop in potential over the interval. Always subtract the smaller V from the larger V when finding the magnitude.
3. $\Delta r = 0.50 - 0.20 = 0.30 \text{ m}$ — [M1] Correct substitution of both values with units.
4. $E = \frac{270}{0.30} = 900 \text{ V m}^{-1}$ — [A1] Correct answer with correct units. V m⁻¹ and N C⁻¹ are equivalent — either is acceptable.
5. $900 \times 0.30 = 270 \text{ V}$ ✓ — Multiplying E back by Δr recovers ΔV = 270 V, confirming the arithmetic is consistent.

Answer: $E = 900 \text{ V m}^{-1}$ (or equivalently 900 N C⁻¹)

Worked example: Finding ΔV from the Area Under an E-r Graph

An E-r graph shows that E = 600 V m⁻¹ at r = 0.10 m and E = 150 V m⁻¹ at r = 0.20 m. The graph is approximately linear between these two points. Estimate the change in electric potential ΔV between r = 0.10 m and r = 0.20 m using the area of a trapezium. [3 marks]

1. $\Delta V = \text{area under } E\text{-}r \text{ graph}$ — [M1] Correct principle stated. Since the graph is approximately linear between the two points, the area is a trapezium.
2. $\text{Area} = \frac{1}{2}(E_1 + E_2) \times \Delta r$ — The two parallel sides of the trapezium are the E values at each end; the width is Δr.
3. $E_1 = 600 \text{ V m}^{-1},\quad E_2 = 150 \text{ V m}^{-1},\quad \Delta r = 0.20 - 0.10 = 0.10 \text{ m}$ — [M1] All values correctly identified with units.
4. $\Delta V = \frac{1}{2}(600 + 150) \times 0.10 = \frac{1}{2} \times 750 \times 0.10$ — Expanding step by step: 600 + 150 = 750; then 750 × 0.10 = 75; then ½ × 75 = 37.5.
5. $\Delta V = 37.5 \text{ V}$ — [A1] Units: V m⁻¹ × m = V. Always check unit consistency at this stage.

Answer: $\Delta V = 37.5 \text{ V}$

6. Comparing E-r and V-r: A Deeper Look

Earlier we established the graph shapes separately. Now connect them: because $V = kQ/r$, differentiating with respect to $r$ gives $dV/dr = -kQ/r^2$. The magnitude of this is $kQ/r^2$, which is exactly $E$. So the $E$-$r$ graph is, in a precise mathematical sense, the gradient function of the $V$-$r$ graph. This is why the $E$-$r$ curve falls off faster — it represents the rate of change of something that is already falling.

A useful comparison for a point charge: at any distance $r$, the ratio $V/E = r$. This means $V = Er$, which is consistent with $V = kQ/r$ and $E = kQ/r^2$ since $(kQ/r^2) \times r = kQ/r$. Note that this relationship $V/E = r$ applies only to a point charge — it does not hold for uniform fields or other charge distributions. Checking this ratio is a quick way to verify your graph readings are consistent with each other when working with point charges.

7. Exam Technique and Common Mistakes

These three spec points appear regularly in AQA papers, and there are predictable ways students lose marks. The guidance below is drawn from mark scheme patterns — use it to write answers that examiners reward.

To consolidate everything: the three spec requirements form a tight triangle of ideas. The graph shapes (1/r² for E, 1/r for V) tell you what to expect visually. The relationship $E = \Delta V / \Delta r$ connects the two quantities analytically and graphically — gradient of V-r gives E. And the area rule ($\Delta V$ = area under E-r graph) is the inverse operation, letting you recover potential changes from field data. Master all three directions and you have complete command of this topic.