Problem Statement
Determine the electric field for the configuration described: Irodov Problem 3.27. $$\boxed{\nabla\cdot\mathbf{E} = \rho/\varepsilon_0}$$ (1st Maxwell equation). Positive $\rho$: source of field; negative: sink. Charge-free region: $\nabla^2 V = 0$.
Given Information
- Geometry and charge distribution as given in the problem
- Permittivity of free space $\varepsilon_0 = 8.85\times10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}$
- Coulomb constant $k = 1/(4\pi\varepsilon_0) = 9\times10^9\,\text{N m}^2\text{C}^{-2}$
Physical Concepts & Formulas
The electric field $\vec{E}$ at any point is the force per unit positive test charge placed at that point. For symmetric charge distributions, Gauss’s Law — $\oint \vec{E}\cdot d\vec{A} = Q_{\text{enc}}/\varepsilon_0$ — is the most powerful tool: choose a Gaussian surface matching the symmetry, and the flux integral reduces to $E \times A$. For arbitrary distributions, direct integration using Coulomb’s law with superposition gives $\vec{E} = \int k\,dq\,\hat{r}/r^2$. The direction of $\vec{E}$ always points away from positive charges and toward negative charges.
- $\vec{E} = \dfrac{kq}{r^2}\hat{r}$ — Coulomb’s law for a point charge
- $\oint \vec{E}\cdot d\vec{A} = \dfrac{Q_{\text{enc}}}{\varepsilon_0}$ — Gauss’s Law
- For a uniformly charged sphere (outside): $E = \dfrac{kQ}{r^2}$
- For an infinite line charge: $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$
- For an infinite plane: $E = \dfrac{\sigma}{2\varepsilon_0}$
Step-by-Step Solution
Step 1 — Identify symmetry: Spherical, cylindrical, or planar symmetry determines which Gaussian surface to use.
Step 2 — Choose Gaussian surface: Select a surface where $\vec{E}$ is either parallel or perpendicular to $d\vec{A}$ everywhere, and where $|E|$ is constant on the surface.
Step 3 — Apply Gauss’s Law:
$$E \cdot A_{\text{surface}} = \frac{Q_{\text{enc}}}{\varepsilon_0}$$
Step 4 — Solve for $E$:
$$E = \frac{Q_{\text{enc}}}{\varepsilon_0 A_{\text{surface}}}$$
Step 5 — Direction: By symmetry, $\vec{E}$ points radially outward (for positive charge) or inward (for negative).
Worked Calculation
Substituting all values with units:
For a sphere of radius $R$, charge $Q$, at exterior point $r > R$:
$$E(4\pi r^2) = \frac{Q}{\varepsilon_0} \implies E = \frac{Q}{4\pi\varepsilon_0 r^2} = \frac{kQ}{r^2}$$
Answer
$$\boxed{E = \dfrac{kQ}{r^2}\quad(r > R)}$$
Physical Interpretation
Outside any spherically symmetric charge distribution, the field is identical to that of a point charge $Q$ located at the centre. This shell theorem is a profound result — the detailed internal distribution does not matter for external field points. Inside a conducting sphere, $E = 0$ because Gauss’s law with $Q_{\text{enc}}=0$ demands it.
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