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Formulas/physics/Electric Charges Fields/Superposition of Electric Fields

Superposition of Electric Fields

Net field at a point due to a system of charges is the vector sum of fields due to individual charges.
Class 11Class 12
Derivation

From force superposition to field superposition

Let charges q1,q2,…,qnq_1, q_2, \ldots, q_n be located at distances r1,r2,…,rnr_1, r_2, \ldots, r_n from point P.

Force on test charge q0q_0 at P:

F⃗=∑i=1nF⃗i=q0∑i=1nqi4πε0ri2r^i\vec{F} = \sum_{i=1}^n \vec{F}_i = q_0 \sum_{i=1}^n \frac{q_i}{4\pi\varepsilon_0 r_i^2}\hat{r}_i

Dividing by q0q_0:

E⃗=F⃗q0=∑i=1nE⃗i\vec{E} = \frac{\vec{F}}{q_0} = \sum_{i=1}^n \vec{E}_i

where E⃗i=qi4πε0ri2r^i\vec{E}_i = \dfrac{q_i}{4\pi\varepsilon_0 r_i^2}\hat{r}_i is the field due to qiq_i alone.

E⃗=14πε0∑i=1nqiri2r^i\boxed{\vec{E} = \frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{n} \frac{q_i}{r_i^2}\hat{r}_i}

Continuous distribution

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

E⃗(r⃗)=14πε0∫ρ(r⃗′)(r⃗−r⃗′)∣r⃗−r⃗′∣3 dV′\vec{E}(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\vec{r}')(\vec{r}-\vec{r}')}{|\vec{r}-\vec{r}'|^3}\, dV'

Similarly for surface (σ\sigma) and line (λ\lambda) distributions by replacing ρ dV′\rho\, dV' with σ dA′\sigma\, dA' or λ dl′\lambda\, dl'.

Note
Vector addition must be done component-wise. Resolve each $\vec{E}_i$ into $x$ and $y$ (and $z$) components, sum separately, then find the resultant magnitude and direction.