Equipotential Surfaces and Radial Fields Revision Notes | A Level AQA Physics

1. The Landscape of Electric Potential

Picture a positive point charge sitting in space. It creates an electric field that pushes other positive charges away — but it also creates something less obvious: a map of electric potential spreading out in all directions. Understanding that map is what this lesson is about.

You already know that electric field lines show the direction a positive test charge would move. Now we add a second layer: equipotential surfaces, which show where the potential is the same. Together, these two pictures give a complete description of the field around any charge.

2. Equipotential Surfaces — What They Are and Why They Matter

Think of a contour map used in geography. Lines of equal height (contours) never cross, and walking along a contour means you neither climb nor descend — you do no work against gravity. Equipotential surfaces in an electric field work exactly the same way: they are the electrical equivalent of contour lines.

For a single point charge, the potential depends only on the distance $r$ from the charge. Every point at the same distance $r$ has the same potential, so the equipotential surfaces are concentric spheres centred on the charge. The closer the sphere, the higher the potential (for a positive charge).

There is a crucial geometric rule: equipotential surfaces are always perpendicular to electric field lines. This is not a coincidence — it follows directly from the definition. If a surface were not perpendicular to the field, there would be a component of the field along the surface, which would mean a potential difference along it, contradicting the definition of an equipotential.

The chart above illustrates how potential decreases as distance increases — the equipotential surfaces become more widely spaced as you move away from the charge, reflecting the weakening field. Closely spaced equipotentials indicate a strong field; widely spaced ones indicate a weak field.

3. No Work Done Along an Equipotential

Here is the key physical argument. Work done moving a charge $q$ between two points is given by $W = q\Delta V$, where $\Delta V$ is the potential difference between those points. On an equipotential surface, $\Delta V = 0$ by definition, so $W = q \times 0 = 0$.

This is not an approximation — it is exact. No matter how far you move the charge along the surface, as long as you stay on the equipotential, the electric field does zero work. The charge's kinetic energy does not change due to the electric force alone.

You can also see this from the field-line geometry. The electric force on a charge acts along the field lines. Moving perpendicular to the field lines (i.e., along an equipotential) means the displacement is always at 90° to the force. Since $W = Fd\cos\theta$ and $\theta = 90°$, we get $\cos 90° = 0$, confirming $W = 0$.

This principle has a direct consequence: to move a charge between two different equipotential surfaces, work must be done — either by the field (if moving to lower potential for a positive charge) or against the field (if moving to higher potential). The amount of work depends on the potential difference between the two surfaces, not on the path taken between them.

4. Electric Potential in a Radial Field

Now we need a formula that tells us the actual value of the potential at any point in the field around a point charge. This is where the radial field formula comes in.

Notice that $V$ is a scalar quantity — it has magnitude and sign, but no direction. This makes it much easier to work with than the electric field $E$, which is a vector. When you have multiple charges, you can find the total potential at a point by adding the individual potentials algebraically, taking signs into account.

The sign of $V$ follows the sign of $Q$. A positive charge creates a positive potential everywhere around it; a negative charge creates a negative potential. Potential is defined as zero at infinity, so as $r \to \infty$, $V \to 0$ — this is consistent with the formula since $\frac{Q}{r} \to 0$.

5. How Potential Varies with Distance — The $V \propto \frac{1}{r}$ Relationship

The formula $V = \frac{Q}{4\pi\varepsilon_0 r}$ tells us that potential is inversely proportional to distance: $V \propto \frac{1}{r}$. Double the distance, halve the potential. This is a softer fall-off than the electric field strength, which follows $E \propto \frac{1}{r^2}$.

On a graph of $V$ against $r$, the curve is a rectangular hyperbola — it falls steeply close to the charge and flattens out as $r$ increases, approaching zero asymptotically. On a graph of $V$ against $\frac{1}{r}$, the relationship is a straight line through the origin with gradient $\frac{Q}{4\pi\varepsilon_0}$.

Comparing the two field quantities: $E \propto \frac{1}{r^2}$ (field strength) and $V \propto \frac{1}{r}$ (potential). The field strength is actually the negative of the potential gradient: $E = -\frac{dV}{dr}$. This relationship is on the AQA specification and connects the two quantities mathematically — differentiating $V = \frac{Q}{4\pi\varepsilon_0 r}$ with respect to $r$ gives $\frac{dV}{dr} = -\frac{Q}{4\pi\varepsilon_0 r^2}$, so $E = \frac{Q}{4\pi\varepsilon_0 r^2}$, which is Coulomb's law for field strength.

6. Worked Calculation — Finding Potential at a Point

Exam questions on this formula are very direct: you are given $Q$ and $r$, and asked to find $V$. The main traps are unit errors and sign errors. The worked example below walks through the method carefully.

Worked example: Calculating Electric Potential in a Radial Field

A small sphere carries a charge of $Q = +4.0 \times 10^{-8}$ C. Calculate the electric potential at a point P that is 0.30 m from the centre of the sphere. Give your answer to 2 significant figures. ($\varepsilon_0 = 8.85 \times 10^{-12}$ F m$^{-1}$)

1. $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$ — [M1] Selecting the correct formula is the first method mark. State it explicitly before substituting.
2. $\frac{1}{4\pi\varepsilon_0} = \frac{1}{4 \times \pi \times 8.85 \times 10^{-12}}$ $= \frac{1}{1.113 \times 10^{-10}} = 8.99 \times 10^{9}$ N m$^2$ C$^{-2}$ — You can use $k = 8.99 \times 10^9$ N m$^2$ C$^{-2}$ directly if given in the data sheet. AQA provides $\varepsilon_0$, so you may need to compute $k$ yourself.
3. $V = 8.99 \times 10^{9} \times \frac{4.0 \times 10^{-8}}{0.30}$ — [M1] Substituting $Q = 4.0 \times 10^{-8}$ C and $r = 0.30$ m with correct units.
4. $\frac{4.0 \times 10^{-8}}{0.30} = \frac{4.0}{0.30} \times 10^{-8} = 13.33... \times 10^{-8} = 1.333 \times 10^{-7}$ C m$^{-1}$ — Keeping an extra significant figure here avoids rounding errors in the final step.
5. $V = 8.99 \times 10^{9} \times 1.333 \times 10^{-7}$ $= 8.99 \times 1.333 \times 10^{9-7}$ $= 11.99 \times 10^{2}$ $= 1199$ V — $8.99 \times 1.333$: compute $9 \times 1.333 = 11.997$, subtract $0.01 \times 1.333 = 0.01333$, giving $11.997 - 0.013 \approx 11.984$. So $V \approx 1198$ V.
6. $V \approx 1200$ V (2 s.f.) — [A1] The answer is positive because $Q$ is positive. Always state the sign explicitly — a negative charge would give a negative potential.

Answer: $V \approx 1200$ V (positive, since $Q > 0$)

Notice that the answer is positive. For a negative source charge, $Q$ would carry a negative sign, making $V$ negative at all finite distances. The magnitude of the calculation is identical — only the sign changes.

7. Connecting Equipotentials, Field Lines, and the Potential Formula

Now that you have the formula, you can calculate the radius of any equipotential surface around a point charge. Rearranging $V = \frac{Q}{4\pi\varepsilon_0 r}$ gives $r = \frac{Q}{4\pi\varepsilon_0 V}$. For a fixed charge $Q$, each value of $V$ corresponds to a specific sphere of radius $r$.

As you move further from the charge, the equipotential spheres become more widely spaced for equal steps in $V$. This is because $V \propto \frac{1}{r}$, so equal decrements in $V$ correspond to increasingly large increments in $r$. The widening spacing of equipotentials is a visual indicator that the field is getting weaker.

The field lines always cross the equipotential spheres at right angles — they point radially outward (for a positive charge) and are therefore perpendicular to the spherical surfaces. This perpendicularity is a universal rule: it holds for any field geometry, not just radial fields.

8. Exam Technique and Common Mistakes

The three spec requirements for this topic — equipotential surfaces, no work done along them, and the potential formula — each have specific exam traps. The guidance below consolidates the most mark-losing errors seen in AQA mark schemes.

For 'describe' questions about equipotential surfaces, examiners want: the shape (concentric spheres for a point charge), the spacing pattern (more widely spaced further from the charge), and the perpendicularity to field lines. Hit all three points for full marks.

With the formula, the geometry, and the work argument all connected, you now have a complete picture of electric potential in a radial field. The three spec requirements — equipotential surfaces, zero work along them, and $V = \frac{Q}{4\pi\varepsilon_0 r}$ — are not isolated facts; they are three views of the same underlying physics.