Electrical Charges and Fields — Formula Sheet

F = k \frac{q_1 q_2}{r^2}

k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2

\vec{F}_{12} = k \frac{q_1 q_2}{r^2} \hat{r}_{12}

E = \frac{F}{q_0}

E = k \frac{q}{r^2}

p = q \cdot 2l

E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0} \frac{2p}{r^3}

E_{\text{equatorial}} = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3}

\phi_E = \vec{E} \cdot \vec{A} = EA \cos\theta

\phi_E = \frac{q_{\text{enc}}}{\varepsilon_0}

E = \frac{\sigma_l}{2\pi\varepsilon_0 r}

E = \frac{\sigma}{\varepsilon_0}

Coulomb’s Law

F ∝ q₁q₂/r²

k = 9 × 10⁹ Nm²/C² in vacuum

Electric field E = F/q₀ (vector, unit: N/C)

For a point charge

E = kq/r²

Dipole field

axial E = 2kp/r³, equatorial E = kp/r³

The electric field at a point is the force experienced per unit positive test charge placed at that point: E = F/q₀.