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Formulas/physics/Electrostatic Potential/Potential due to Point Charge

Potential due to Point Charge

Potential at distance r from a point charge q. Positive for positive charge, negative for negative charge. Zero at infinity.
Class 11Class 12
Derivation

Setup

Place point charge qq at the origin. The field at distance rr:

E⃗=q4πε0r2r^\vec{E} = \frac{q}{4\pi\varepsilon_0 r^2}\hat{r}

Integration along a radial path

Since E⃗\vec{E} is radial, choose the path of integration along a radial line (any path gives the same result — the field is conservative):

V(r)=−∫∞rE⃗⋅dr⃗′=−∫∞rq4πε0r′2 dr′V(r) = -\int_{\infty}^{r} \vec{E} \cdot d\vec{r}' = -\int_{\infty}^{r} \frac{q}{4\pi\varepsilon_0 r'^2}\, dr' =−q4πε0[−1r′]∞r=−q4πε0(−1r+0)= -\frac{q}{4\pi\varepsilon_0}\left[-\frac{1}{r'}\right]_{\infty}^{r} = -\frac{q}{4\pi\varepsilon_0}\left(-\frac{1}{r} + 0\right) V=q4πε0r\boxed{V = \frac{q}{4\pi\varepsilon_0 r}}

Sign and behaviour

  • q>0q > 0: V>0V > 0, decreasing from +∞+\infty at the charge to 00 at infinity
  • q<0q < 0: V<0V < 0, increasing from −∞-\infty at the charge to 00 at infinity
  • V∝1/rV \propto 1/r, falls off slower than E∝1/r2E \propto 1/r^2
Note
Potential from a point charge is spherically symmetric — it depends only on distance $r$, not on direction. Surfaces of constant $V$ are concentric spheres centred on $q$.