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

Electric Potential: The Point‑Charge Formula

The Point‑Charge Formula

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:

• \(Q\) = charge in coulombs (C)
• \(r\) = distance from the charge in metres (m)
• \(\varepsilon_0 = 8.85\times10^{-12}\,\text{F/m}\) (vacuum permittivity)

Deriving the Formula (Quick Sketch)

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.

Worked Example

🔍 Find the potential at a point 0.5 m from a charge of +3 µC.

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.

Units & Conversion

• Potential \(V\) is measured in volts (V).
• 1 V = 1 J/C (joule per coulomb).
• When using SI units, always keep \(Q\) in coulombs, \(r\) in metres, and \(\varepsilon_0\) in farads per metre.

Common Mistakes to Avoid

• Mixing up units: µC vs C, m vs cm.
• Forgetting the \(4\pi\) factor.
• Using the wrong sign for negative charges (the formula gives a negative potential).
• Assuming the potential is the same as the electric field—remember, potential is a scalar, field is a vector.

Summary