1B · Fields by Gauss Law — Sphere, Cylinder, Zero-Point + Solved PYQs

Using Gauss's law, find the field of a uniformly charged solid sphere outside and inside. Show the graphical variation and prove the field is zero inside (at the centre).

Recall

Steps

1. State the symmetry: $\vec{E} = E(r)\hat{r}$, so on a concentric Gaussian sphere, $\Phi = E(r)\cdot 4\pi r^2$.
2. Outside ($r > R$): enclosed charge is the full $Q$; Gauss's law gives $E = Q/4\pi\varepsilon_0 r^2$ (point-charge form).
3. Inside ($r < R$): enclosed charge scales with volume, $Q_{\text{in}} = Q\,r^3/R^3$; Gauss's law gives $E = Q\,r/4\pi\varepsilon_0 R^3$ (linear in $r$).
4. At the centre ($r = 0$): Eq. (2) gives $E = 0$. For a thin shell, $Q_{\text{in}} = 0$ everywhere inside, so $E = 0$ throughout.
5. Graph: straight line from origin to $(R, E_{\max})$, then a $1/r^2$ falling curve — as in Fig 1.1.

Answer

Exam tip

Calculate the electric field on the surface of a $^{238}_{92}$U nucleus (radius $7\times 10^{-15}$ m). Take $\varepsilon_0 = 8.85\times 10^{-12}$ C²/N·m², $e = 1.6\times 10^{-19}$ C.

Recall

Steps

1. Charge of the nucleus: $Q = 92\,e = 92\times 1.6\times 10^{-19} = 1.472\times 10^{-17}$ C (reason: 92 protons).
2. At the surface $r = R$: $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R^2}$ with $1/4\pi\varepsilon_0 = 9\times 10^9$ N·m²/C².
3. Numerator: $9\times 10^9\times 1.472\times 10^{-17} = 1.3248\times 10^{-7}$ N·m²/C.
4. Denominator: $R^2 = (7\times 10^{-15})^2 = 49\times 10^{-30}$ m².
5. Divide: $E = \dfrac{1.3248\times 10^{-7}}{49\times 10^{-30}} = 2.7\times 10^{21}$ N/C, radially outward (reason: positive charge pushes a test charge away).

Answer

Exam tip

Using Gauss's theorem, find the electric field intensity at a point due to a charged cylinder (long line of charge).

Recall

Steps

1. Let charge per unit length be $\lambda$. Draw a coaxial Gaussian cylinder of radius $r$ and length $l$.
2. Flux through the curved face only: $\Phi = E\cdot 2\pi r l$ (reason: $\vec{E}\parallel d\vec{S}$ on curved face; on caps $\vec{E}\perp d\vec{S}$, so they contribute 0).
3. Enclosed charge: $Q_{\text{in}} = \lambda l$. Apply Gauss's law: $E\cdot 2\pi r l = \lambda l/\varepsilon_0$.
4. Cancel $l$: $E = \lambda/2\pi\varepsilon_0 r$, radially outward (reason: charge is positive).

Answer

Exam tip

Two charges $8\ \mu$C and $2\ \mu$C are placed 50 cm apart in air. At which point is the electric field intensity zero?

Recall

Steps

1. Let the point be at distance $x$ from $8\ \mu$C along the line joining the charges, so it is $(50 - x)$ cm from $2\ \mu$C. Between them the two fields point in opposite directions (reason: each charge pushes a test charge away from itself).
2. Set the magnitudes equal: $\dfrac{k\cdot 8}{x^2} = \dfrac{k\cdot 2}{(50-x)^2}$. Cancel $k$: $\dfrac{8}{x^2} = \dfrac{2}{(50-x)^2}$.
3. Take square roots (cleaner than expanding squares): $\dfrac{\sqrt{8}}{x} = \dfrac{\sqrt{2}}{50-x}$ → $\dfrac{2\sqrt{2}}{x} = \dfrac{\sqrt{2}}{50-x}$ → $\dfrac{2}{x} = \dfrac{1}{50-x}$.
4. Cross-multiply: $2(50 - x) = x$ → $100 - 2x = x$ → $3x = 100$ → $x = 33.3$ cm.
5. Check: from $2\ \mu$C, the distance is $50 - 33.3 = 16.7$ cm. Verify: $8/(33.3)^2 \approx 2/(16.7)^2 \approx 0.0072$. ✓

Answer

Exam tip