Electric Potential: Definition and Work Done Revision Notes | A Level AQA Physics

1. Why Electric Potential Matters

You already know that charges exert forces on each other. But to calculate how much energy is involved when a charge moves through a field, you need a new quantity: electric potential.

Think back to gravitational potential energy. When you lift a mass, you do work against gravity, and that energy is stored. Electric potential works on exactly the same principle — except the field is electric, not gravitational, and the charges involved can be positive or negative.

The key question this lesson answers is: how much energy is transferred when a charge moves from one point to another in an electric field? The answer comes from two tightly linked ideas — absolute electric potential and the work done equation $\Delta W = Q\Delta V$.

2. Absolute Electric Potential: Definition and Zero at Infinity

Picture a single positive point charge sitting in space. It creates an electric field that spreads outward in all directions. Now imagine bringing a small positive test charge from very far away — effectively from infinity — towards it. You have to do work against the repulsive force the whole way in. That work done per unit charge is the absolute electric potential at the point where you stop.

The choice of zero at infinity is a convention, but it is a physically meaningful one. At infinity, the field has no influence on the test charge — no force acts and no work is done. Every other point in the field has a potential measured relative to that baseline.

Near a positive source charge, the potential is positive — you had to do positive work to push the test charge in against repulsion. Near a negative source charge, the potential is negative — the field actually does the work for you as the test charge is attracted inward, so the potential energy decreases below the infinity baseline.

This sign behaviour is one of the most important features of electric potential at A-level. The sign tells you about the nature of the source charge and the direction of energy change when a positive test charge moves to that point.

3. Electric Potential Difference

In most exam questions, you are not moving a charge from infinity — you are moving it between two specific points, A and B, that are both already inside the field. What matters then is the electric potential difference $\Delta V$ between those two points.

Notice that potential difference is a scalar quantity — it has magnitude and sign, but no direction. The sign of $\Delta V$ tells you whether the potential increases or decreases as you move from A to B.

This connects directly to the absolute potential idea from the previous section. Because absolute potential is defined relative to infinity, you can always find $\Delta V$ between any two points by subtracting their absolute potentials. The reference point (infinity) cancels out in the subtraction, which is precisely why defining zero at infinity is so useful.

4. Work Done Moving a Charge: The Core Equation

Now that potential difference is defined, the energy calculation follows directly. When a charge $Q$ moves through a potential difference $\Delta V$, the work done is given by a single, elegant equation.

The sign of $\Delta W$ carries physical meaning. A positive $\Delta W$ means work is done on the charge — energy is transferred into the charge (its electric potential energy increases). A negative $\Delta W$ means the field does work on the charge — the charge loses potential energy and gains kinetic energy, or energy is released to the surroundings.

This sign logic is where many students lose marks. The sign of $\Delta W$ depends on both the sign of $Q$ and the sign of $\Delta V$. You must treat both carefully rather than working with magnitudes alone.

5. Worked Examples Using ΔW = QΔV

Theory is only useful when you can apply it under exam conditions. The following worked examples cover the two scenarios you are most likely to encounter: moving a positive charge and moving a negative charge, each with a clear sign analysis.

Worked example: Work Done Moving a Positive Charge Between Two Points

A proton (charge $+1.6 \times 10^{-19}$ C) is moved from point A, where the electric potential is $+400$ V, to point B, where the electric potential is $+100$ V. Calculate the work done on the proton. [3 marks]

1. $\Delta W = Q \Delta V$ — State the equation first — this earns the method mark even if arithmetic goes wrong later.
2. $\Delta V = V_{\text{final}} - V_{\text{initial}} = V_B - V_A = 100 - 400 = -300$ V — Always subtract initial from final. Here the potential decreases, so $\Delta V$ is negative.
3. $\Delta W = (+1.6 \times 10^{-19}) \times (-300)$ — Carry the sign of Q explicitly. The proton has a positive charge.
4. $\Delta W = 1.6 \times 10^{-19} \times 300 = 4.8 \times 10^{-17}$ J, with a negative sign: $\Delta W = -4.8 \times 10^{-17}$ J — Arithmetic check: $1.6 \times 300 = 480$, so $1.6 \times 10^{-19} \times 300 = 4.8 \times 10^{-17}$. The negative result means the proton loses electric potential energy moving from A to B — consistent with it moving to a lower potential.

Answer: $\Delta W = -4.8 \times 10^{-17}$ J (the proton loses electric potential energy)

Notice how the sign of the answer tells a physical story. The proton moved from higher to lower potential — the same direction a positive charge naturally tends to move in an electric field. The field did the work, so the proton's potential energy decreased (negative $\Delta W$).

Worked example: Work Done Moving a Negative Charge Between Two Points

An electron (charge $-1.6 \times 10^{-19}$ C) is moved from point X, where the electric potential is $-200$ V, to point Y, where the electric potential is $+600$ V. Calculate the work done on the electron. [3 marks]

1. $\Delta W = Q \Delta V$ — State the equation to secure the method mark.
2. $\Delta V = V_Y - V_X = +600 - (-200) = +800$ V — Take care with the double negative: subtracting $-200$ gives $+200$, so $\Delta V = 600 + 200 = 800$ V.
3. $\Delta W = (-1.6 \times 10^{-19}) \times (+800)$ — The electron's charge is negative — carry this sign through.
4. $1.6 \times 10^{-19} \times 800 = 1.28 \times 10^{-16}$ J, with negative sign: $\Delta W = -1.28 \times 10^{-16}$ J — Arithmetic check: $1.6 \times 800 = 1280$, so $1.6 \times 10^{-19} \times 800 = 1.28 \times 10^{-16}$. The negative result means the electron loses electric potential energy. The electron naturally moves toward higher potential because the field does work on it — no external agent is required to drive this motion. The field drives the electron from X to Y, releasing energy in the process.

Answer: $\Delta W = -1.28 \times 10^{-16}$ J (the electron loses electric potential energy moving to higher potential)

6. Interpreting Sign: A Systematic Approach

Sign errors are the single most common source of lost marks on electric potential questions. Rather than guessing, use a consistent two-step check every time.

• Step 1 — Find $\Delta V$: always compute $V_{\text{final}} - V_{\text{initial}}$, keeping all signs.
• Step 2 — Multiply by $Q$: use the signed value of $Q$ (positive for protons/positive ions, negative for electrons/negative ions).
• Step 3 — Interpret: positive $\Delta W$ means energy is transferred into the charge (external agent does work); negative $\Delta W$ means the charge loses potential energy (field does the work or energy is released).

A useful cross-check: a positive charge naturally moves from high to low potential (like water flowing downhill). If you are moving a positive charge in that natural direction, $\Delta W$ should be negative — the field does the work, not you. If you are pushing it uphill (low to high potential), $\Delta W$ is positive — you supply the energy.

For negative charges, the logic reverses. A negative charge naturally moves from low to high potential (attracted toward positive regions). Moving it in that direction gives negative $\Delta W$; moving it against that direction gives positive $\Delta W$.

7. Exam Technique and Common Mistakes

Electric potential questions reward precision in language and sign handling. The following guidance targets the specific patterns that distinguish full-mark responses from partial ones at A-level.

For extended-response questions asking you to 'explain' work done in terms of potential, structure your answer as: state the potential difference between the two points, identify the sign and magnitude of Q, apply $\Delta W = Q\Delta V$, then interpret the sign of the result in terms of energy transfer direction. This four-part chain reliably captures all available marks.

These two spec requirements — the definition of absolute electric potential with zero at infinity, and the work done equation $\Delta W = Q\Delta V$ — form the foundation for all further work on electric fields, including equipotential surfaces and the relationship between field strength and potential gradient. Mastering the sign conventions here pays dividends across the entire fields topic.