Problem Statement
Determine the electric field for the configuration described: A thin spherical shell of radius $R$ carries total surface charge $q$. Find $E$ inside and outside. See problem statement for all given quantities. Gauss’s law relates the electric flux through any closed surface to the total enclosed charge. It is one of Maxwell’s four equations and is especially p
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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