Gauss's law — find electric field due to uniformly charged sphere (inside and outside)

Question

A solid non-conducting sphere of radius $R$ carries a total charge $Q$ distributed uniformly throughout its volume. Using Gauss’s law, find the electric field at a point (a) outside the sphere ( $r > R$ ) and (b) inside the sphere ( $r < R$ ).

(JEE Main 2023, similar pattern)

Solution — Step by Step

The charge distribution has spherical symmetry. So the electric field must be radial and depend only on $r$ . We choose a concentric spherical Gaussian surface of radius $r$ .

Gauss’s law: $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}$

Since $\vec{E}$ is radial and constant on the Gaussian sphere: $E \cdot 4\pi r^2 = \frac{Q_{enc}}{\varepsilon_0}$

The Gaussian surface encloses the entire charge: $Q_{enc} = Q$.

E \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0}

\boxed{E = \frac{Q}{4\pi\varepsilon_0 r^2} = \frac{kQ}{r^2} \quad (r > R)}

Outside, the sphere behaves exactly like a point charge at its centre.

The Gaussian surface of radius $r$ encloses only a fraction of the total charge. Volume charge density: $\rho = \frac{Q}{\frac{4}{3}\pi R^3}$ .

Enclosed charge: $Q_{enc} = \rho \cdot \frac{4}{3}\pi r^3 = Q \cdot \frac{r^3}{R^3}$

E \cdot 4\pi r^2 = \frac{Q r^3}{\varepsilon_0 R^3}

\boxed{E = \frac{Qr}{4\pi\varepsilon_0 R^3} = \frac{kQr}{R^3} \quad (r < R)}

Inside, the field increases linearly with $r$ .

Outside: $E(R) = kQ/R^2$ . Inside: $E(R) = kQR/R^3 = kQ/R^2$ . Both match at the surface — the field is continuous, confirming our results.

Why This Works

Gauss’s law works beautifully here because of the spherical symmetry. Symmetry tells us that $\vec{E}$ must be radial, so the flux integral simplifies to $E \times 4\pi r^2$ . The physics is entirely in the enclosed charge.

Inside the sphere, only the charge within radius $r$ contributes to the field — the outer shell contributes zero net field (this is the shell theorem). This is why the field grows linearly inside: more enclosed charge as $r$ increases.

Alternative Method

For the outside field, you can directly use Coulomb’s law and integrate over the sphere. But that requires a triple integral that gives the same result. Gauss’s law provides the answer in two lines — this is its power for symmetric charge distributions.

The graph of $E$ vs $r$ is a classic exam question. The field increases linearly from 0 at the centre to $kQ/R^2$ at the surface, then falls as $1/r^2$ outside. Sketch this graph with labels — it is worth 2-3 marks in CBSE boards.

Common Mistake

For the inside case, students often write $Q_{enc} = Q$ instead of $Q \cdot r^3/R^3$ . Inside the sphere, you only enclose the charge within your Gaussian surface, not the total charge. The enclosed charge scales as $r^3$ (volume ratio), not $r^2$ . This error gives $E \propto 1/r^2$ inside, which is wrong — the correct result is $E \propto r$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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