use V = Q / (4πε0r) for the electric potential in the field due to a point charge

Electric Potential – Point Charge

In the Cambridge A‑Level Physics syllabus (9702) the concept of electric potential is central to understanding how charges interact at a distance. For a single point charge the potential at a distance r from the charge is given by

$$V = \frac{Q}{4\pi\varepsilon_{0}r}$$

where:

• V – electric potential (volts, V)
• Q – magnitude of the point charge (coulombs, C)
• r – radial distance from the charge centre (metres, m)
• \varepsilon_{0} – permittivity of free space, $ \varepsilon_{0}=8.854\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$

Derivation from the Electric Field

The electric field of a point charge is

$$E = \frac{Q}{4\pi\varepsilon_{0}r^{2}}$$

Since the electric potential difference between two points A and B is the negative line integral of the field,

$$V_B - V_A = -\int_{A}^{B}\mathbf{E}\cdot d\mathbf{l}$$

Choosing A at infinity where the potential is zero and B at distance r, the integral becomes

$$V(r) = -\int_{\infty}^{r}\frac{Q}{4\pi\varepsilon_{0}r^{2}}\,dr = \frac{Q}{4\pi\varepsilon_{0}r}$$

Using the Formula

1. Identify the charge Q (include sign – positive charge gives positive potential, negative gives negative).
2. Measure or be given the distance r from the charge centre.
3. Insert the values into $V = \dfrac{Q}{4\pi\varepsilon_{0}r}$.
4. Check units: ensure Q is in coulombs and r in metres; the result will be in volts.

Worked Example

Problem: A point charge of $+5.0\ \mu\text{C}$ is located at the origin. Calculate the electric potential 0.20 m from the charge.

Solution:

$$ \begin{aligned} Q &= +5.0\times10^{-6}\ \text{C} \\ r &= 0.20\ \text{m} \\ V &= \frac{5.0\times10^{-6}}{4\pi(8.854\times10^{-12})(0.20)} \\ &= \frac{5.0\times10^{-6}}{2.22\times10^{-11}} \\ &\approx 2.25\times10^{5}\ \text{V} \end{aligned} $$

The potential is $+2.3\times10^{5}\ \text{V}$ (rounded to two significant figures).

Common Pitfalls

• Forgetting the sign of Q: a negative charge yields a negative potential.
• Using centimetres or millimetres for r without converting to metres.
• Assuming the potential at the surface of a charged sphere is the same as for a point charge at the centre; this is true only outside the sphere.
• Mixing up electric field $E$ (V m⁻¹) with electric potential $V$ (V).

Summary Table

Quantity	Symbol	Value / Expression	Units
Permittivity of free space	$\varepsilon_{0}$	$8.854\times10^{-12}$	C² N⁻¹ m⁻²
Electric potential due to point charge	$V$	$\displaystyle \frac{Q}{4\pi\varepsilon_{0}r}$	V
Electric field due to point charge	$E$	$\displaystyle \frac{Q}{4\pi\varepsilon_{0}r^{2}}$	V m⁻¹

Practice Questions

1. Calculate the potential at 0.10 m from a charge of $-2.0\ \mu\text{C}$.
2. A proton is placed 5.0 cm from a point charge of $+3.0\times10^{-9}\ \text{C}$. Determine the potential energy of the proton at that position.
3. Two identical point charges $+Q$ are separated by 0.30 m. Find the potential midway between them.

These notes provide the essential framework for applying $V = \dfrac{Q}{4\pi\varepsilon_{0}r}$ in A‑Level examinations. Mastery of the sign conventions, unit handling, and the relationship between field and potential will enable you to solve a wide range of problems involving point charges.