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Gauss's law with symmetry

Spherical, cylindrical and planar symmetry, nonuniform densities, slabs, nested shells and conductors, all with $E \times \text{area} = Q_{\text{enc}}/\varepsilon_0$.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to use Gauss's law to find fields of symmetric distributions, including nonuniform densities, slabs and nested shells, and to check the results at boundaries.

2. What you already have

You know Gauss's law, $\oint\vec{E} \cdot d\vec{a} = Q_{\text{enc}}/\varepsilon_0$, its differential form, and the fields of uniformly charged spheres, lines and planes from Physics C. You can integrate in spherical and cylindrical coordinates. This lesson makes Gauss's law a reliable tool for harder cases: densities that vary, slabs, and several charged objects at once.

3. Words for this lesson

4. Flux, symmetry and one line of algebra

Gauss's law, $\oint\vec{E} \cdot d\vec{a} = Q_{\text{enc}}/\varepsilon_0$, is always true, but it gives the field only when symmetry lets $E$ come out of the integral. That happens for three families:

• Spherical symmetry: $\vec{E} = E(r)\hat{r}$; a concentric sphere gives $E \cdot 4\pi r^2$.
• Cylindrical symmetry: $\vec{E} = E(s)\hat{s}$; a coaxial cylinder of length $l$ gives $E \cdot 2\pi sl$.
• Planar symmetry: $\vec{E} = E(x)\hat{x}$; a pillbox of face area $A$ gives $2EA$ if the charge is symmetric about the middle.

In each case the field follows from

$$E \times (\text{area of the surface}) = \frac{Q_{\text{enc}}}{\varepsilon_0}.$$

When the density varies, the enclosed charge needs an integral. For a ball with $\rho = \rho_0r/R$, the charge inside radius $r$ is $\int_0^r\rho_0\frac{r'}{R}4\pi r'^2\,dr' = \pi\rho_0r^4/R$, so $E = \rho_0r^2/(4\varepsilon_0R)$. For several objects, add the charges each surface encloses. Outside any spherically symmetric distribution, the field is $kQ_{\text{total}}/r^2$; inside a hollow symmetric shell, the shell contributes nothing.

5. The method, step by step, and how to check it

1. Argue the symmetry first. State why the field must be radial, or perpendicular to a plane, and why its size depends on only one coordinate. Without that argument Gauss's law gives only the flux.
2. Pass the surface through the field point. The radius of the sphere or cylinder is the distance at which you want $E$.
3. Integrate for the enclosed charge whenever $\rho$ is not uniform, using $d\tau = 4\pi r^2dr$ for spheres and $2\pi sl\,ds$ for cylinders.
4. Solve piece by piece in regions separated by boundaries: inside a ball, between shells, outside everything.

Checks. The field must be continuous across any boundary that carries no surface charge, and jump by $\sigma/\varepsilon_0$ across one that does. Outside a finite distribution it must fall as $kQ/r^2$ with the total charge. At the center of a symmetric distribution it must be zero. And $\nabla \cdot \vec{E}$ computed from your answer must return $\rho/\varepsilon_0$.

6. Why charges outside do not spoil the argument

Charges outside a Gaussian surface contribute zero net flux: every field line from them that enters the surface also leaves it. But they do contribute to the field at each point of the surface. The reason Gauss's law still gives the field in symmetric problems is that symmetry constrains the total field, not just the part from the enclosed charge.

Inside a spherical shell, for instance, the shell's own field is not obviously zero: the near side is closer but smaller in apparent size, the far side larger but farther away. Newton proved for gravity that these effects cancel exactly, and Gauss's law with spherical symmetry gives the same result in one line. Put the shell off center relative to other charges, or dent it, and the symmetry is gone; Gauss's law remains true but no longer tells you the field.

7. Slabs and sheets

For a slab of thickness $2d$ with uniform density $\rho$, a pillbox extending from $-x$ to $x$ encloses $\rho\,2xA$, and both faces carry outward flux $EA$. Inside, $E = \rho x/\varepsilon_0$, growing linearly from zero at the middle plane. Outside, the pillbox encloses the whole thickness, $\rho\,2dA$, so $E = \rho d/\varepsilon_0 = \sigma/2\varepsilon_0$ with $\sigma = 2\rho d$ the charge per unit area: the same as an infinitely thin sheet.

Semiconductor devices use exactly this geometry. In the depletion region of a p–n junction, ionized dopants form two slabs of opposite charge, and Gauss's law gives a field that rises linearly through one slab and falls through the other, peaking at the junction. The peak field and the width of the region set the diode's breakdown voltage, and engineers calculate them this way.

8. Nonuniform densities

Real charge distributions are rarely uniform. The electron cloud of a hydrogen atom has density $\propto e^{-2r/a}$; a charged dielectric sphere may have charge concentrated near its surface. Gauss's law still works as long as the density depends only on $r$: compute $Q_{\text{enc}}(r) = \int_0^r\rho(r')4\pi r'^2dr'$ and divide by $4\pi\varepsilon_0r^2$.

The shape of $E(r)$ then reflects where the charge sits. A density growing outward gives a field growing faster than linearly inside; a density concentrated at the center gives a field that rises quickly and then falls as $1/r^2$ even while still inside the distribution. Geophysicists use the same reasoning with gravity: the way $g$ varies with depth inside Earth reveals that the planet's density increases sharply toward the iron core.

9. Charge on conductors

In a conductor in equilibrium the field inside is zero, or free charges would move. By Gauss's law, any surface drawn inside the metal encloses no net charge, so all excess charge lives on the surface. Just outside, a pillbox with one face inside the metal gives $E = \sigma/\varepsilon_0$, perpendicular to the surface.

A charged conducting sphere therefore has $E = kQ/R^2$ at its surface, and the surface field limits how much charge it can hold in air. A small sphere reaches breakdown with little charge; a large one holds far more. That is why Van de Graaff generators have big, smooth domes, and why sharp points, where the local radius of curvature is small and the field concentrated, leak charge through corona discharge — the principle of the lightning rod.

10. When Gauss's law is not enough

A finite cylinder, a cube, a disk, two separate spheres: none has a surface on which the field is constant, so Gauss's law cannot be solved for $E$. It still holds, and it still tells you useful things — the total flux, the absence of charge in empty regions — but the field itself must come from integrating Coulomb's law or from solving Laplace's equation.

Approximate symmetry is often good enough. The field near the middle of a long wire, far from its ends, is nearly that of an infinite line; the field near the center of a large capacitor plate is nearly that of an infinite sheet. Knowing when such approximations hold — typically within a few percent when the distance is less than a tenth of the object's size — is part of using Gauss's law well.

11. Gauss's law for gravity

Newton's law of gravitation has the same inverse-square form as Coulomb's law, so gravity obeys its own Gauss's law, $\oint\vec{g} \cdot d\vec{a} = -4\pi GM_{\text{enc}}$. Everything in this lesson carries over: the field inside a hollow planet would be zero; inside a uniform planet $g$ grows linearly with radius; outside any spherical body, the body acts as a point mass.

The analogy has practical consequences. Gravimeters measure tiny variations in $g$ at Earth's surface, and geologists and oil companies use Gauss's-law reasoning to infer the density of rock below. NASA's GRACE satellites, flying in tandem and measuring the distance between them to micrometers, have mapped month-by-month changes in Earth's gravity caused by melting ice sheets and depleted aquifers, including the groundwater losses in California's Central Valley.

12. In the world: Van de Graaff generators

Robert Van de Graaff built his first electrostatic generator at Princeton in 1929: a moving belt carries charge up into a hollow metal sphere, where it flows to the outer surface because the field inside a conductor is zero. Charge can be carried in indefinitely, and the sphere's potential climbs until the field at its surface, $kQ/R^2$, reaches the breakdown strength of the surrounding gas.

That limit sets the design. A science-museum dome $0.25$ m in radius tops out near $750$ kV and about $21$ μC — enough to make hair stand on end and throw sparks half a meter long. Research accelerators built on the principle, such as the tandem Van de Graaffs used for decades in U.S. nuclear physics laboratories, enclose the terminal in a tank of pressurized sulfur hexafluoride, whose breakdown field is several times higher than air's, and reach potentials above $20$ million volts.

13. In the world: depletion regions in diodes

Every diode, solar cell and transistor contains p–n junctions. Where p-type and n-type silicon meet, mobile electrons and holes diffuse across and annihilate, leaving behind a region of fixed ionized dopants: negative on the p side, positive on the n side. The two charged layers are slabs, and Gauss's law gives the field inside them exactly as in this lesson: rising linearly through one slab, peaking at the junction, falling through the other.

With doping of $10^{16}$ atoms per cubic centimeter, the charge density is about $1.6 \times 10^{3}$ C/m³, and a depletion region a fraction of a micrometer wide produces peak fields of several million volts per meter. In a solar cell, this built-in field sweeps apart the electrons and holes created by sunlight, producing the current. Power-device engineers design the slabs' widths and densities with Gauss's law so that their devices block hundreds of volts without breaking down.

14. Gauss's law is always true, but it gives the field only with symmetry

A common misreading is that the field at a point on a Gaussian surface depends only on the charge inside. The flux depends only on the enclosed charge; the field at each point depends on every charge, inside and outside. Only when symmetry makes the field constant over the surface does the flux determine the field.

The opposite error is to think Gauss's law fails when charge lies outside. It never fails; outside charges add zero net flux. Inside a symmetric shell, for instance, the shell's charge is all outside any smaller sphere, and the field there is exactly zero. What fails without symmetry is only the step from flux to field.

15. The field inside and outside a uniform ball

1. Choose a sphere of radius $r < R$.

   $\oint\vec{E} \cdot d\vec{a} = 4\pi r^2E$

   Radial and uniform on the sphere.

2. Find the enclosed charge.

   $Q_{\text{enc}} = Q\dfrac{r^3}{R^3}$

   The fraction of the volume inside.

3. Solve for the inside field.

   $E = \dfrac{kQr}{R^3}$

   Linear in $r$.

4. Repeat for $r > R$.

   $4\pi r^2E = \dfrac{Q}{\varepsilon_0} \quad\Rightarrow\quad E = \dfrac{kQ}{r^2}$

   All the charge is enclosed.

5. Check continuity at the surface.

   $\dfrac{kQR}{R^3} = \dfrac{kQ}{R^2}$

   No surface charge, so no jump.

16. A long cylinder with a nonuniform density

1. A long cylinder of radius $a$ has $\rho = \beta s$. Choose a coaxial cylinder of radius $s < a$ and length $l$.

   $\oint\vec{E} \cdot d\vec{a} = E \cdot 2\pi sl$

   No flux through the flat ends.

2. Integrate for the enclosed charge.

   $Q_{\text{enc}} = \displaystyle\int_0^s\beta s' \cdot 2\pi s'l\,ds' = \dfrac{2\pi\beta ls^3}{3}$

   Cylindrical shells of volume $2\pi s'l\,ds'$.

3. Apply Gauss's law.

   $E \cdot 2\pi sl = \dfrac{2\pi\beta ls^3}{3\varepsilon_0}$

   Flux equals enclosed charge over $\varepsilon_0$.

4. Solve for the inside field.

   $E = \dfrac{\beta s^2}{3\varepsilon_0}$

   The length cancels, as it must.

5. Find the outside field.

   $s > a: \quad E = \dfrac{\beta a^3}{3\varepsilon_0s}$

   Falls as $1/s$, like a line charge with $\lambda = 2\pi\beta a^3/3$.

6. Check with the divergence.

   $\dfrac{1}{s}\dfrac{d}{ds}\left(s \cdot \dfrac{\beta s^2}{3\varepsilon_0}\right) = \dfrac{\beta s}{\varepsilon_0} = \dfrac{\rho}{\varepsilon_0}$

   The differential form agrees.

17. A charged ball inside a charged shell

1. A uniform ball of radius $R$ holds $+Q$; a concentric shell of radius $2R$ holds $-2Q$. Find the field for $r < R$.

   $E = \dfrac{kQr}{R^3}$

   Only part of the ball is enclosed; the shell is outside.

2. Find the field between the ball and the shell.

   $R < r < 2R: \quad E = \dfrac{kQ}{r^2}$

   The whole ball is enclosed; the shell is not.

3. Find the charge enclosed outside the shell.

   $r > 2R: \quad Q_{\text{enc}} = Q - 2Q = -Q$

   Both objects are inside.

4. Find the field outside.

   $E = -\dfrac{kQ}{r^2}$

   Pointing inward.

5. Check the jump at the shell.

   $E(2R^+) - E(2R^-) = -\dfrac{kQ}{4R^2} - \dfrac{kQ}{4R^2} = -\dfrac{2kQ}{4R^2}$

   Comparing just outside and just inside.

6. Compare with the surface charge.

   $\dfrac{\sigma}{\varepsilon_0} = \dfrac{-2Q}{4\pi(2R)^2\varepsilon_0} = -\dfrac{2kQ}{4R^2}$

   The jump equals $\sigma/\varepsilon_0$, as it must.

7. Find where the field is largest.

   $E_{\max} = E(R) = \dfrac{kQ}{R^2}$

   At the ball's surface.

8. Sketch the result.

   $E: \ \nearrow \text{ to } R, \ \searrow \text{ to } 2R, \ \text{jumps negative}, \ \to 0$

   Rising, falling, flipping, fading.

18. Your turn: a uniform ball has $kQ/R^2 = 600$ N/C at its surface. What is the field at $r = R/3$?

1. Write the inside field.

   $E = \dfrac{kQ}{R^3}r = \dfrac{kQ}{R^2}\cdot\dfrac{r}{R}$

   Linear in $r$.

2. Substitute the values.

   $E = 600 \times \tfrac{1}{3}$

   A third of the radius.

3. Your turn: work this step out. Its working is at the end of the packet.

   Evaluate the field.

19. Guided practice

A thin spherical shell of radius $R$ carries $7$ nC spread uniformly over it. What is the field at a point inside, at $r = R/2$?

• zero
• $kQ/(R/2)^2$, as if the charge were at the center
• $kQ/R^2$, the same as at the surface
• impossible to find, since charge outside the surface affects the field

20. Guided practice

Complete the worked solution: a uniform ball of radius $R$ holds $312$ nC, and a concentric thin shell of radius $2R$ holds $7$ nC. Find the charge enclosed by Gaussian spheres of radius $R/2$ and $3R$, and their difference, in nC.

1. Find the charge inside radius $R/2$.

   $Q(R/2) = 312 \times \left(\tfrac{1}{2}\right)^3 =$ _p

   An eighth of the ball's volume.

2. Find the charge inside radius $3R$.

   $Q(3R) = 312 + 7 =$ _q

   The whole ball and the whole shell.

3. Subtract the two enclosed charges.

   $Q(3R) - Q(R/2) =$ _r

   The charge lying between the two surfaces.

21. Guided practice

Match each charge distribution to the Gaussian surface that finds its field.

22. Practice

A uniformly charged ball of radius $R$ has $kQ/R^2 = 769$ N/C at its surface. Fill in the field, in N/C, at $r = R/2$, $r = R$ and $r = 2R$.

23. Practice

A thick, wide slab carries a uniform charge density $\rho = 4$ μC/m³. What is the field inside the slab, $5$ cm from its middle plane, in N/C? Use $\varepsilon_0 = 8.85 \times 10^{-12}$ C²/(N m²).

Answer: N/C

24. Practice

A ball of radius $R = 8$ cm has charge density $\rho = \rho_0r/R$ with $\rho_0 = 4$ μC/m³. What is the electric field at $r = 6$ cm, in N/C? Use $\varepsilon_0 = 8.85 \times 10^{-12}$ C²/(N m²).

Answer: N/C

25. Somewhere new

A Van de Graaff generator in a science museum has a metal sphere of radius $0.25$ m. Air breaks down at about $3.0 \times 10^6$ V/m. What is the most charge the sphere can hold, in μC? Use $k = 8.99 \times 10^9$ N m²/C².

Answer: μC

26. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

27. Test question

A uniformly charged ball of radius $R$ has $kQ/R^2 = 372$ N/C at its surface. Fill in the field, in N/C, at $r = R/2$, $r = R$ and $r = 2R$.

28. What you can do now

You can find fields with Gauss's law. Explain to someone why the field inside a charged spherical shell is zero.

Working for the steps left to you

18. Your turn: a uniform ball has $kQ/R^2 = 600$ N/C at its surface. What is the field at $r = R/3$?, step 3