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Formulas/physics/Electrostatic Potential/Superposition of Potentials

Superposition of Potentials

Net potential at a point due to a system of charges is the algebraic (scalar) sum of individual potentials. No direction involved — simpler than field superposition.
Class 11Class 12
Derivation

From field superposition

The electric field satisfies superposition: E⃗=∑iE⃗i\vec{E} = \sum_i \vec{E}_i.

The potential is defined via the line integral of E⃗\vec{E}:

V=−∫∞PE⃗⋅dr⃗=−∫∞P∑iE⃗i⋅dr⃗=∑i(−∫∞PE⃗i⋅dr⃗)=∑iViV = -\int_{\infty}^{P} \vec{E} \cdot d\vec{r} = -\int_{\infty}^{P} \sum_i \vec{E}_i \cdot d\vec{r} = \sum_i \left(-\int_{\infty}^{P} \vec{E}_i \cdot d\vec{r}\right) = \sum_i V_i

Result

V=∑i=1nqi4πε0ri\boxed{V = \sum_{i=1}^{n} \frac{q_i}{4\pi\varepsilon_0 r_i}}

where rir_i is the distance from charge qiq_i to the field point.

Scalar vs vector superposition

This is an algebraic sum — signs of charges matter, but no direction is involved. Compare with field superposition which requires vector addition of nn quantities.

For a continuous distribution with volume charge density ρ(r⃗′)\rho(\vec{r}'):

V(r⃗)=14πε0∫ρ(r⃗′)∣r⃗−r⃗′∣ dV′V(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\vec{r}')}{|\vec{r} - \vec{r}'|}\, dV'
Remember
Potential superposition is strictly simpler than field superposition. For problems where only the potential (not the field direction) is needed, always work with $V$ rather than $\vec{E}$.