Electric potential

CIE A-Level Physics· 9702 2025-2027 Syllabus 21.2 Electric potential· 15 min read

1. Electric Potential and Potential Energy★★☆☆☆⏱ 4 min

📘 Definition

Electric potential

V

The work done per unit positive charge to bring a small test charge from infinity to the point in question. Units are volts (V), where 1 V = 1 J C⁻¹, and V = 0 at infinity.

Example:

Potential is a scalar quantity, so only magnitude (and sign) matters, no direction.

Change in electric potential energy $Δ E_{p}$ of a charge $q$ moving through a potential difference $Δ V$ is given by:

\Delta E_p = q \Delta V

📐 Worked Example

A +2.0 μC charge is moved from a point at 100 V to a point at 500 V. Calculate the change in electric potential energy.

1
Identify known values: $q = 2.0 \times 1 0^{- 6}$ C, $Δ V = V_{f ina l} - V_{ini t ia l} = 500 - 100 = 400$ V
2
Substitute into the potential energy relation:
3
$\Delta E_p = (2.0 \times 10^{-6}) \times 400 = 8.0 \times 10^{-4} \text{ J}$
4
A positive charge moving to higher potential gains potential energy, matching our positive result.

2. Potential in Common Charge Configurations★★★☆☆⏱ 6 min

For a point charge $Q$ , the electric potential at distance $r$ from the charge is derived from Coulomb's law:

V = \frac{1}{4\pi \epsilon_0} \frac{Q}{r}

Since potential is scalar, the total potential at a point near multiple charges is just the algebraic sum of individual potentials, no vector addition needed. For parallel plates with separation $d$ and total potential difference $V$ , potential at distance $x$ from the 0 V plate is $V (x) = \frac{V x}{d}$ .

📐 Worked Example

Calculate the total electric potential at point P, 0.5 m from a +2 nC charge and 1.0 m from a -3 nC charge. Use $\frac{1}{4 π ϵ _{0}} = 9.0 \times 1 0^{9}$ N m² C⁻².

1
Calculate potential from the first charge:
2
$V_1 = \frac{9.0 \times 10^9 \times 2.0 \times 10^{-9}}{0.5} = 36 \text{ V}$
3
Calculate potential from the second charge:
4
$V_2 = \frac{9.0 \times 10^9 \times (-3.0) \times 10^{-9}}{1.0} = -27 \text{ V}$
5
Sum the potentials to get total:
6
$V_{total} = 36 + (-27) = 9 \text{ V}$

3. Electric Field as Potential Gradient★★★☆☆⏱ 5 min

📘 Definition

Potential gradient

The rate of change of electric potential with distance in the direction of the electric field.

Electric field strength is equal to the negative of the potential gradient:

E = -\frac{dV}{dr}

For uniform electric fields between parallel plates, this simplifies to $E = \frac{V}{d}$ , where $V$ is potential difference across plates separated by distance $d$ . The negative sign confirms electric field always points from higher potential to lower potential.

📐 Worked Example

Potential along the x-axis is given by $V = 3 x^{2} - 2 x$ , where $V$ is in V and $x$ in m. Calculate electric field strength at $x = 2$ m.

1
Use $E = - \frac{d V}{d x}$ , first differentiate V:
2
$\frac{dV}{dx} = 6x - 2$
3
Substitute into the E relation:
4
$E = -(6x - 2)$
5
Insert $x = 2$ m:
6
$E = -(6(2) - 2) = -10 \text{ N C}^{-1}$
7
The negative sign means E points in the negative x-direction, magnitude 10 N C⁻¹.

4. Equipotential Surfaces★★☆☆☆⏱ 3 min

An equipotential surface is any surface where every point has the same electric potential. No work is done moving a charge along an equipotential, because $Δ V = 0$ so $Δ E_{p} = 0$ .

Equipotential surfaces are always perpendicular to electric field lines
For a point charge, equipotentials are concentric spheres centered on the charge
Between parallel plates, equipotentials are equally spaced parallel planes
Closer equipotentials correspond to stronger electric fields

📐 Worked Example

Explain why electric field lines must always cross equipotential surfaces at 90°.

1
If a field line crossed at an angle other than 90°, there would be a component of the electric field parallel to the equipotential surface.
2
A parallel component of E means there is a potential gradient along the surface, so potential would not be constant everywhere on the surface, contradicting the definition of an equipotential.

5. Common Pitfalls

Wrong move:

Treating electric potential as a vector and adding potentials with direction components

Why:

Potential is a scalar quantity, only electric field is a vector. Incorrect vector addition leads to wrong total potential.

Correct move:

Add all potential values algebraically, keeping their signs (positive for positive charges, negative for negative charges), no direction adjustment needed.

Wrong move:

Forgetting the negative sign in $E = - d V / d r$ and getting the direction of E wrong

Why:

Examiners regularly test understanding of the relationship between potential direction and electric field direction.

Correct move:

Always include the negative sign, and confirm that E always points from higher potential to lower potential to check your result.

Wrong move:

Confusing electric potential with electric potential energy

Why:

The terms are closely related but have different definitions, units, and physical meanings.

Correct move:

Remember that potential V is work done per unit charge, while potential energy $Δ E_{p} = q Δ V$ is the total energy for a specific charge q.

Wrong move:

Using the point charge potential formula inside a uniformly charged conducting sphere

Why:

The formula $V = Q / (4 π ϵ_{0} r)$ only applies outside the sphere; E = 0 inside a conductor.

Correct move:

For any point inside a conducting charged sphere, potential is constant and equal to the potential at the surface of the sphere.

6. Quick Reference Cheatsheet

Concept	Formula	Key Notes
Electric potential definition	$V = W / q$	Scalar, $V = 0$ at infinity
Potential from point charge	$V = \frac{Q}{4 π ϵ _{0} r}$	Sum algebraically for multiple charges
E as potential gradient	$E = - \frac{d V}{d r}$	$E = V / d$ for uniform fields
Potential energy change	$Δ E_{p} = q Δ V$	Units: J, not V
Equipotential property	$E ⊥$ equipotential	No work done along equipotential

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 22
Point charge potential calculation
2023 · 13
Potential gradient vs electric field
2024 · 21
Equipotential surface properties

Going deeper

syllabusOfficial CIE 9702 2025 SyllabusFull syllabus reference

What's Next

Electric potential is a foundational concept for nearly all subsequent electricity topics in CIE A-Level Physics. It is used to derive capacitance, calculate energy stored in capacitors, and analyse the motion of charged particles through potential differences. Mastery of potential and its relation to electric field will also help you understand concepts in modern physics like particle acceleration and electric potential in atomic systems. Many exam questions combine electric potential with other electrostatics concepts, so solid understanding here pays dividends across the paper.