Field of a uniformly charged sphere

Consider a sphere of radius $R$ with total charge $Q$ distributed uniformly through the volume, and let us find the electric field at distance $r$ from the centre. Spherical symmetry guarantees that the field is radial and of constant magnitude on every concentric sphere: we can therefore choose as the Gaussian surface a sphere of radius $r$ and take $E$ outside the flux integral, $\Phi = E\cdot 4\pi r^2$.

Outside the sphere ($r > R$). The Gaussian surface encloses all the charge $Q$. Gauss’s theorem gives:

$E \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0} \quad\Rightarrow\quad E = k\,\frac{Q}{r^2}$

Outside, the sphere behaves exactly like a point charge concentrated at the centre: the field decreases as $1/r^2$.

Inside the sphere ($r < R$). Now the Gaussian surface encloses only the fraction of charge contained in the sphere of radius $r$. Since the density is uniform, the enclosed charge scales with the volume, that is as $r^3/R^3$:

$Q_{\text{int}} = Q\,\frac{r^3}{R^3} \quad\Rightarrow\quad E \cdot 4\pi r^2 = \frac{Q_{\text{int}}}{\varepsilon_0} \quad\Rightarrow\quad E = k\,\frac{Q}{R^3}\,r$

Inside, the field is thus linear in $r$: it starts from zero at the centre, grows to a maximum at the surface ($r = R$), and from there decreases as $1/r^2$.

Key formula

Uniformly charged sphere of radius $R$ and charge $Q$: $\ev{\begin{aligned} E &= k\,\frac{Q}{r^2} & &(r > R) \\ E &= k\,\frac{Q}{R^3}\,r & &(r < R) \end{aligned}}$ The field grows linearly inside, reaches its maximum at $r = R$, then decays as $1/r^2$ outside.

Collegamenti

Argomenti: Campo elettrico e potenziale Concetti: Teorema di Gauss Competenze: Applicazione del teorema di Gauss Metodi: Teorema di Gauss Oggetti: Sfera carica

Esercizi collegati: Which graph of E · From the field to the spherical radius · Speculative physics: Coulomb 1 over r cubed