⚡️ Electric potential (V) tells us how much potential energy a unit charge would have at a point in an electric field. Think of it like the height of a hill: the higher the hill, the more potential energy a ball would have if placed there.
For a single point charge \(Q\) the electric potential at a distance \(r\) is given by:
\$V = \frac{Q}{4\pi\varepsilon_0 r}\$
Where:
We start from Coulomb’s law for the electric field of a point charge:
\$E = \frac{Q}{4\pi\varepsilon_0 r^2}\$
Electric potential is the negative integral of the electric field from a reference point (usually infinity) to the point of interest:
\$V = -\int{\infty}^{r} E\,dr = -\int{\infty}^{r} \frac{Q}{4\pi\varepsilon_0 r^2}\,dr\$
Carrying out the integral gives:
\$V = \frac{Q}{4\pi\varepsilon_0 r}\$
??
The negative sign disappears because the field points away from a positive charge and we’re integrating from infinity to a finite point.
🔍 Find the potential at a point 0.5 m from a charge of +3 µC.
| Parameter | Value |
|---|---|
| \(Q\) | \(3 \times 10^{-6}\,\text{C}\) |
| \(r\) | \(0.5\,\text{m}\) |
| \(\varepsilon_0\) | \(8.85 \times 10^{-12}\,\text{F/m}\) |
Plugging into the formula:
\$V = \frac{3 \times 10^{-6}}{4\pi (8.85 \times 10^{-12}) (0.5)} \approx 2.7 \times 10^{5}\,\text{V}\$
So the potential is about 270 kV.
When you see a problem asking for potential at a distance from a point charge, remember:
📝 Write down the formula first, then substitute numbers. This keeps your work clear and reduces mistakes.
⚡️ The electric potential due to a point charge is a simple yet powerful tool. By remembering the formula \(V = Q/(4\pi\varepsilon_0 r)\) and practicing unit conversions, you can tackle most potential questions in the Cambridge A‑Level Physics exam.
💡 Keep your calculations neat, double‑check units, and you’ll ace the potential problems with confidence!