Chapter 1 · Electric Charges and Fields

1.3 Electric Flux, Gauss's Law and Its Applications

Count the field lines that leave a closed surface and you have counted the charge inside: $\oint\mathbf E\cdot d\mathbf A=q_\text{enc}/\varepsilon_0$. With symmetry, this one law gives the fields of a long wire, a large sheet and a spherical shell in a single line each.

Car passengers caught in a thunderstorm are remarkably safe. A lightning strike may dump several coulombs onto the car body, yet the people inside feel nothing: the charge stays on the outside of the metal shell and the field inside is almost exactly zero. The same idea protects sensitive electronics inside metal boxes, MRI rooms lined with copper mesh, and the passengers of every aircraft that is struck by lightning (on average, each airliner is hit about once a year). Why is the field inside a closed charged shell zero, however much charge it carries?

Coulomb's law and superposition answer such questions in principle, but adding up the fields of millions of charge elements is hard work. In this section we first describe smoothly spread-out charge by its charge density. Then we define the electric flux through a surface — roughly, the number of field lines crossing it — and meet Gauss's law, which says that the flux out of any closed surface depends only on the charge inside it. Whenever the charge is arranged with enough symmetry, Gauss's law turns a difficult sum into a one-line calculation. We use it to find the fields of an infinite line of charge, an infinite plane sheet and a thin spherical shell, and to explain the safety of the car in the storm.

Continuous Charge Distributions

A charged rod, a charged metal plate or a charged balloon carries billions of electrons spread over its length, surface or volume. At the scale of centimetres we can ignore the graininess (Section 1.1) and treat the charge as smeared out smoothly, just as we treat water as a continuous fluid rather than as molecules.

📘 Definition 1.6 — Linear, surface and volume charge density
If a small element of a charged body carries charge $\Delta q$, the charge density there is:
  • linear charge density $\lambda=\dfrac{\Delta q}{\Delta l}$ for charge spread along a line or thin wire (unit $\text{C/m}$);
  • surface charge density $\sigma=\dfrac{\Delta q}{\Delta A}$ for charge spread over a surface (unit $\text{C/m}^2$);
  • volume charge density $\rho=\dfrac{\Delta q}{\Delta V}$ for charge spread through a volume (unit $\text{C/m}^3$).
Here $\Delta l$, $\Delta A$ and $\Delta V$ are small on the everyday scale but still contain very many elementary charges. A distribution is uniform when the density is the same everywhere: then simply $\lambda=Q/L$, $\sigma=Q/A$ or $\rho=Q/V$.

To find the field of a continuous distribution, cut it into small elements, treat each element $dq$ ($=\lambda\,dl$, $\sigma\,dA$ or $\rho\,dV$) as a point charge with field $d\mathbf E=k\,\dfrac{dq}{r^2}\,\hat{\mathbf r}$, and add all the $d\mathbf E$ vectors (uses integration): $$\mathbf E=\int k\,\frac{dq}{r^2}\,\hat{\mathbf r}.$$ This is superposition again, with the sum replaced by an integral. It is easy only when symmetry makes most of the components cancel, as in the next example.

✏️ Example 1 — Field on the axis of a charged ring
A thin ring of radius $R$ carries a charge $Q$ spread uniformly round it. (a) Show that at a point $P$ on its axis, a distance $x$ from its centre, the field is $E=\dfrac{kQx}{(x^2+R^2)^{3/2}}$ along the axis. (b) Evaluate it for $Q=+3.0$ nC, $R=4.0$ cm, $x=3.0$ cm. (c) What happens at the centre and very far away?
Symmetry Every element $dq$ of the ring is the same distance $r=\sqrt{x^2+R^2}$ from $P$. Its field $k\,dq/r^2$ has a component along the axis and a component perpendicular to it. The element on the opposite side of the ring has an equal and opposite perpendicular component, so all perpendicular components cancel in pairs.
(a) Add Each axial component is $k\dfrac{dq}{r^2}\cos\alpha$ with $\cos\alpha=\dfrac xr$, the same for every element. So $$E=\frac{kx}{r^3}\sum dq=\frac{kQx}{(x^2+R^2)^{3/2}},$$ pointing away from the ring along the axis if $Q\gt0$.
(b) Numbers $x^2+R^2=(0.030)^2+(0.040)^2=2.5\times10^{-3}\ \text{m}^2$, so $(x^2+R^2)^{3/2}=1.25\times10^{-4}\ \text{m}^3$ and $$E=\frac{(9.0\times10^9)(3.0\times10^{-9})(0.030)}{1.25\times10^{-4}}\approx6.5\times10^{3}\ \text{N/C}.$$
(c) Limits At the centre ($x=0$), $E=0$: the elements pull equally in all directions. For $x\gg R$, $E\approx kQ/x^2$: from far away the ring looks like a point charge. ✓

Electric Flux

Imagine the field lines as a steady flow and hold a small wire frame in it. How many lines pass through the frame? That depends on three things: how strong the field is (how dense the lines are), how large the frame is, and how it is tilted — face-on to the lines it catches the most, edge-on it catches none. To keep track of the tilt we give each flat area a direction.

📘 Definition 1.7 — Area vector and electric flux
The area vector of a small flat surface is $\Delta\mathbf A=\Delta A\,\hat{\mathbf n}$, where $\hat{\mathbf n}$ is a unit vector normal (perpendicular) to the surface. For a closed surface — one that encloses a volume, like a sphere or a box — $\hat{\mathbf n}$ always points outwards.
The electric flux through the surface element is $\Delta\Phi_E=\mathbf E\cdot\Delta\mathbf A=E\,\Delta A\cos\theta$, where $\theta$ is the angle between $\mathbf E$ and $\hat{\mathbf n}$. The flux through a whole surface is the sum over all its elements: $$\Phi_E=\int\mathbf E\cdot d\mathbf A,$$ written $\oint\mathbf E\cdot d\mathbf A$ for a closed surface. For a flat surface of area $A$ in a uniform field, $\Phi_E=EA\cos\theta$. Flux is a scalar; its SI unit is $\text{N m}^2/\text{C}$.

For a closed surface, flux is positive where field lines leave (outward, $\theta\lt90^\circ$) and negative where they enter ($\theta\gt90^\circ$). The net flux out of a closed surface is therefore, roughly, "lines out minus lines in". A uniform field gives zero net flux through any closed surface: every line that enters also leaves.

✏️ Example 2 — Flux through a tilted square, and through a closed cube
A uniform field $\mathbf E=2.0\times10^3\,\hat{\mathbf i}$ N/C. (a) Find the flux through a square of side $10$ cm when its normal makes $0^\circ$, $60^\circ$ and $90^\circ$ with $\mathbf E$. (b) A cube of side $10$ cm has its edges along the axes. Find the flux through each face and the net flux out of the cube.
Formula Flat surface, uniform field: $\Phi=EA\cos\theta$ with $A=(0.10)^2=1.0\times10^{-2}\ \text{m}^2$, so $EA=20\ \text{N m}^2/\text{C}$.
(a) Tilts $\theta=0^\circ$: $\Phi=20\ \text{N m}^2/\text{C}$ (face-on). $\theta=60^\circ$: $\Phi=20\cos60^\circ=10\ \text{N m}^2/\text{C}$. $\theta=90^\circ$: $\Phi=0$ — the field lines skim along the surface.
(b) Cube The face whose outward normal is $+\hat{\mathbf i}$: $\Phi=+20$. The face with outward normal $-\hat{\mathbf i}$: $\theta=180^\circ$, $\Phi=-20$. The four other faces are parallel to $\mathbf E$: $\Phi=0$ each.
Net $20-20+0=0$. Twenty units of flux enter on the left and leave on the right; the cube encloses no charge, and Gauss's law below says that this is exactly why the net flux is zero.

Gauss's Law

Start with the simplest case: a point charge $q$ at the centre of a sphere of radius $r$. On the sphere the field has the same magnitude $E=\dfrac{q}{4\pi\varepsilon_0r^2}$ everywhere and points straight outwards, along $\hat{\mathbf n}$. So $\mathbf E\cdot d\mathbf A=E\,dA$ on every element, and $$\oint\mathbf E\cdot d\mathbf A=E\oint dA=\frac{q}{4\pi\varepsilon_0r^2}\cdot4\pi r^2=\frac{q}{\varepsilon_0}.$$ The radius has cancelled: a bigger sphere has more area but a weaker field, in exactly compensating amounts, because $E\propto1/r^2$ while area $\propto r^2$. In the language of field lines, every line from $q$ crosses every sphere around it once, so the count does not depend on the size of the sphere.

Now deform the sphere into any shape — a box, an egg, a potato. Each line from $q$ still has to get out, so the net count of lines crossing outwards stays the same. If a line happens to cross the surface three times (out, in, out), the in and out crossings cancel and it still counts once. And a charge outside the surface sends every line in and out again, contributing zero net flux. Adding up the contributions of many charges by superposition gives one of the most important laws in physics.

📐 Law 1.8 — Gauss's law

The net electric flux out of any closed surface is equal to $1/\varepsilon_0$ times the total charge enclosed by the surface: $$\oint\mathbf E\cdot d\mathbf A=\frac{q_\text{enc}}{\varepsilon_0}.$$ Here $q_\text{enc}$ is the algebraic sum (signs included) of all the charges inside the surface, and $\mathbf E$ is the total field at the surface, due to all charges, inside and outside. The closed surface used is called a Gaussian surface; it is an imaginary surface and need not coincide with any material object.

Notice what Gauss's law does and does not say. Charges outside the surface contribute to $\mathbf E$ at each point of the surface, but their contributions to the net flux cancel. Zero net flux does not mean zero field on the surface — it means as many lines enter as leave. And the law is exact only because of the inverse-square form of Coulomb's law; it is, in fact, Coulomb's law in another form. (It also holds for moving charges, which makes it one of Maxwell's equations — Chapter 8.) A charge lying exactly on the surface is a borderline case to avoid when choosing a Gaussian surface.

✏️ Example 3 — Counting flux with Gauss's law
(a) A charge $q=+2.0\ \mu\text{C}$ sits at the centre of a cube of side $20$ cm. Find the net flux out of the cube and the flux through one face. Does the answer change if the cube is replaced by a sphere of radius $50$ cm? (b) A closed surface encloses $+3.0$ nC and $-5.0$ nC; a charge of $+4.0$ nC lies outside it. Find the net flux out of the surface.
Law $\Phi=q_\text{enc}/\varepsilon_0$, whatever the shape and size of the closed surface.
(a) Cube $$\Phi=\frac{2.0\times10^{-6}}{8.85\times10^{-12}}\approx2.26\times10^{5}\ \text{N m}^2/\text{C}.$$ The charge is at the centre, so by symmetry each of the six faces gets one-sixth: $\Phi_\text{face}\approx3.77\times10^{4}\ \text{N m}^2/\text{C}$. The side length ($20$ cm) is never needed. The sphere encloses the same charge, so its flux is the same, $2.26\times10^5\ \text{N m}^2/\text{C}$.
(b) Mixed $q_\text{enc}=+3.0-5.0=-2.0$ nC (the outside charge does not count): $$\Phi=\frac{-2.0\times10^{-9}}{8.85\times10^{-12}}\approx-226\ \text{N m}^2/\text{C}.$$ Negative: more lines enter than leave.
Check Interactive 1, arrangement “Example 3(b)”, integrates $\mathbf E\cdot d\mathbf A$ numerically over the surface and reads $\approx-226$; the $+4.0$ nC charge contributes $\approx0$. Change the shape or size — the reading stays put until a charge crosses the surface.
Interactive 1 — Gauss's Law Lab: Flux through Any Closed Surface

You see a slice through a closed surface (sphere, cube or long box) and point charges in the slicing plane; grid squares are $1$ cm. The page adds up $\mathbf E\cdot\Delta\mathbf A$ over thousands of small patches of the whole 3-D surface. The ticks round the outline show the outward (red) or inward (blue) component of $\mathbf E$. Drag the charges or the surface, change its shape, or press Play to make it grow and shrink: the flux changes only when a charge crosses the surface.

charge enclosed $q_\text{enc}$—
$\oint\mathbf E\cdot d\mathbf A$, numerical (N m²/C)—
$q_\text{enc}/\varepsilon_0$ (N m²/C)—
flux from each charge—

Applications: Line, Sheet and Shell

Gauss's law is always true, but it gives the field directly only when we can pick a Gaussian surface on which the field is known in direction and constant in size (or perpendicular to the surface, so it adds no flux). Symmetry tells us where such surfaces are.

🧭 Problem-solving strategy — finding a field with Gauss's law
  1. Use symmetry to decide the direction of $\mathbf E$ and on which surfaces its magnitude is constant: radial from a line (cylindrical symmetry), perpendicular to a sheet (planar), radial from a centre (spherical).
  2. Choose a Gaussian surface through the point of interest made of pieces where $\mathbf E$ is either parallel to $d\mathbf A$ with constant magnitude, or perpendicular to $d\mathbf A$ (no flux): a coaxial cylinder, a pillbox straddling the sheet, a concentric sphere.
  3. Write the flux as $E\times(\text{area where }\mathbf E\parallel d\mathbf A)$ and the enclosed charge from the charge density.
  4. Solve $E\times\text{area}=q_\text{enc}/\varepsilon_0$ for $E$, and check the limits and the units.

An infinitely long straight line of charge with uniform linear charge density $\lambda$. By symmetry the field points radially away from the line (for $\lambda\gt0$) and has the same size at every point a distance $r$ from it. Take as Gaussian surface a cylinder of radius $r$ and length $l$ with the line as its axis. On the two flat end caps $\mathbf E$ lies along the surface, so their flux is zero. On the curved surface $\mathbf E$ is along $\hat{\mathbf n}$ and constant, so the flux is $E(2\pi rl)$. The enclosed charge is $\lambda l$: $$E\,(2\pi rl)=\frac{\lambda l}{\varepsilon_0}.$$

📐 Result 1.9 — Field of an infinitely long line charge

At distance $r$ from an infinitely long straight line with uniform linear charge density $\lambda$, $$\mathbf E=\frac{\lambda}{2\pi\varepsilon_0r}\,\hat{\mathbf n}=\frac{2k\lambda}{r}\,\hat{\mathbf n},$$ where $\hat{\mathbf n}$ is the unit vector perpendicular to the line, pointing away from it (the field points towards the line if $\lambda\lt0$). It falls off as $1/r$. For a long but finite wire it is a good approximation at points near the middle whose distance from the wire is small compared with its length.

An infinite plane sheet of charge with uniform surface charge density $\sigma$. By symmetry the field is perpendicular to the sheet, points away from it on both sides (for $\sigma\gt0$), and has the same size at equal distances on either side. Take a cylinder ("pillbox") of cross-section $A$ that pierces the sheet at right angles. The curved side has $\mathbf E$ along its surface: no flux. Each flat end has flux $EA$, outwards. The enclosed charge is $\sigma A$: $$2EA=\frac{\sigma A}{\varepsilon_0}.$$

📐 Result 1.10 — Field of an infinite plane sheet of charge

Near an infinite plane sheet with uniform surface charge density $\sigma$, on either side, $$\mathbf E=\frac{\sigma}{2\varepsilon_0}\,\hat{\mathbf n},$$ where $\hat{\mathbf n}$ is the unit normal pointing away from the sheet on that side. The field is uniform: it does not depend on the distance from the sheet. For a large but finite sheet this holds at points whose distance from it is small compared with its size and that are far from its edges. Two such parallel sheets carrying $+\sigma$ and $-\sigma$ (a parallel-plate arrangement) produce $E=\sigma/\varepsilon_0$ between them, from $+$ to $-$, and $E=0$ outside.

A thin spherical shell of radius $R$ carrying charge $Q$ spread uniformly over its surface, $\sigma=Q/(4\pi R^2)$. By spherical symmetry the field is radial and has the same magnitude at every point a distance $r$ from the centre. Take as Gaussian surface a concentric sphere of radius $r$; its flux is $E(4\pi r^2)$. Outside ($r\gt R$) it encloses all of $Q$, so $E(4\pi r^2)=Q/\varepsilon_0$. Inside ($r\lt R$) it encloses no charge at all, so $E(4\pi r^2)=0$.

📐 Result 1.11 — Field of a uniformly charged thin spherical shell

For a thin shell of radius $R$ carrying charge $Q$ uniformly:

Outside ($r\gt R$): $$\mathbf E=\frac{1}{4\pi\varepsilon_0}\,\frac{Q}{r^2}\,\hat{\mathbf r},$$ exactly as if all the charge were a point charge at the centre. Just outside the surface this is $E=\dfrac{kQ}{R^2}=\dfrac{\sigma}{\varepsilon_0}$.

Inside ($r\lt R$): $$\mathbf E=\mathbf 0.$$

The zero inside is not obvious from Coulomb's law. A point inside the shell near one side is close to a small patch of charge and far from a large patch on the other side; the inverse-square law makes their pulls cancel exactly. This is the physics of the car in the storm: the charge on a metal body spreads over its outer surface, and the field inside is zero. (Conductors are treated fully in Section 2.3.) It is also why Section 1.1 could treat charged spheres as point charges from outside.

✏️ Example 4 — The field of a long charged wire
A long straight wire carries a uniform charge of $+4.0$ nC per metre. (a) Find the field $2.0$ cm from the wire, away from its ends. (b) Find the force on an electron at that point. (c) At what distance is the field $1200$ N/C?
Model Points near the middle of a long wire, close to it: use the infinite-line result $E=2k\lambda/r$ with $\lambda=4.0\times10^{-9}$ C/m.
(a) Field $$E=\frac{2(9.0\times10^9)(4.0\times10^{-9})}{0.020}=3.6\times10^{3}\ \text{N/C},$$ radially outwards from the wire.
(b) Force $F=eE=(1.6\times10^{-19})(3.6\times10^3)\approx5.8\times10^{-16}$ N, towards the wire (the electron is negative).
(c) Distance $E\propto1/r$, and $1200=3600/3$, so $r=3\times2.0=6.0$ cm. (Compare a point charge, whose field would fall to a third at only $\sqrt3$ times the distance.)
Check Interactive 2, “line”, opens on $\lambda=4.0$ nC/m; set $r=2.0$ cm and read $E=3600$ N/C.
✏️ Example 5 — One sheet, and two parallel sheets
Large flat sheets carry a uniform surface charge density of magnitude $\sigma=8.85\times10^{-7}\ \text{C/m}^2$. Find the field (a) near a single positive sheet; (b) between and outside two parallel sheets carrying $+\sigma$ and $-\sigma$; (c) between and outside two parallel sheets both carrying $+\sigma$.
Single sheet $\dfrac{\sigma}{2\varepsilon_0}=\dfrac{8.85\times10^{-7}}{2(8.85\times10^{-12})}=5.0\times10^{4}$ N/C, perpendicular to the sheet and away from it on both sides — at any distance (as long as it is small compared with the sheet).
(b) $+\sigma,-\sigma$ Add the two sheets' fields (superposition). Between them both point from $+$ to $-$: $E=2\times5.0\times10^4=1.0\times10^{5}$ N/C. Outside, they point in opposite directions and cancel: $E=0$.
(c) $+\sigma,+\sigma$ Between them the fields are opposite: $E=0$. Outside, both point away from the pair: $E=1.0\times10^5$ N/C.
Meaning Arrangement (b) is the parallel-plate capacitor of Chapter 2: a uniform field confined between the plates. Interactive 2, “plane sheet”, shows the single-sheet field as a step from $-\sigma/2\varepsilon_0$ to $+\sigma/2\varepsilon_0$.
✏️ Example 6 — Inside and outside a charged shell
A thin spherical shell of radius $10$ cm carries $+5.0$ nC spread uniformly. Find (a) the surface charge density; the field at (b) $5.0$ cm, (c) just outside the surface and (d) $30$ cm from the centre.
(a) Density $\sigma=\dfrac{Q}{4\pi R^2}=\dfrac{5.0\times10^{-9}}{4\pi(0.10)^2}\approx3.98\times10^{-8}\ \text{C/m}^2$.
(b) Inside $r=5.0$ cm $\lt R$: a Gaussian sphere there encloses no charge, so $E=0$.
(c) Surface $E=\dfrac{kQ}{R^2}=\dfrac{(9.0\times10^9)(5.0\times10^{-9})}{(0.10)^2}=4.5\times10^{3}$ N/C, outwards. Check with $\sigma/\varepsilon_0=\dfrac{3.98\times10^{-8}}{8.85\times10^{-12}}\approx4.5\times10^3$ N/C ✓.
(d) Outside $E=\dfrac{kQ}{r^2}=\dfrac{45}{(0.30)^2}=500$ N/C, outwards — a ninth of the surface value at three times the radius.
Check Interactive 2, “spherical shell”, opens on this shell: the graph is zero up to $10$ cm, jumps to $4500$ N/C and then falls as $1/r^2$.
Interactive 2 — Line, Sheet and Shell: Gaussian Surfaces and E against r

Pick a charge distribution. The sketch shows the Gaussian surface that symmetry suggests — a coaxial cylinder, a pillbox or a concentric sphere — passing through the field point at distance $r$; the graph plots $E$ against $r$ and the marker follows the point. Press Play to sweep $r$, or set it with the slider. The readouts apply Gauss's law: flux $=q_\text{enc}/\varepsilon_0$, then $E=$ flux $\div$ area.

4.0
field point—
$q_\text{enc}$—
flux $q_\text{enc}/\varepsilon_0$ (N m²/C)—
$E$ at the point (N/C)—
⚠️ Common mistakes with flux and Gauss's law

1. "$\mathbf E$ in Gauss's law is due only to the enclosed charge." No — $\mathbf E$ is the total field. Outside charges change $\mathbf E$ on the surface but not the net flux.

2. Using Gauss's law without symmetry. The law holds for every closed surface, but it gives $E$ only if $E$ is constant (and along $d\mathbf A$) on the parts that carry flux. For a finite rod or a square plate it is true but useless for finding $E$.

3. Mixing up $\sigma/2\varepsilon_0$ and $\sigma/\varepsilon_0$. One sheet gives $\sigma/2\varepsilon_0$. Two oppositely charged sheets give $\sigma/\varepsilon_0$ between them. The field just outside a charged shell (or a conductor) is $\sigma/\varepsilon_0$, because there is no field on the inner side.

4. Forgetting the sign of the flux. Flux into a closed surface is negative; a surface enclosing only negative charge has negative net flux.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — a uniformly charged solid sphere, and a charge at the corner of a cube

Not required by the CBSE syllabus. Electric flux and Gauss's law are listed by JEE Main, NEET and JEE Advanced. JEE Main and NEET name only the wire, the plane sheet and the thin shell; JEE Advanced lists "simple cases, such as" these, so the solid sphere fits it most directly. No syllabus names the cube-corner case.

Solid sphere. An insulating ball of radius $R$ carries charge $Q$ spread uniformly through its volume, $\rho=\dfrac{Q}{\tfrac43\pi R^3}$. Outside, a concentric Gaussian sphere encloses all of $Q$: $E=kQ/r^2$, as for the shell. Inside, a Gaussian sphere of radius $r$ encloses only $Q\dfrac{r^3}{R^3}$, so $$E\,(4\pi r^2)=\frac{Qr^3}{\varepsilon_0R^3}\ \Rightarrow\ E=\frac{kQ\,r}{R^3}\quad(r\le R).$$ The field grows linearly from zero at the centre to $kQ/R^2$ at the surface, then falls as $1/r^2$ — no jump, unlike the shell.

Charge at a corner. A charge $q$ at one corner of a cube is shared by $8$ cubes that meet there, so this cube receives $\dfrac{q}{8\varepsilon_0}$. Of its six faces, the three that touch the corner contain the charge in their planes and get no flux; the other three share the flux equally: $\dfrac{q}{24\varepsilon_0}$ each. These "build a bigger symmetric surface" tricks are JEE favourites.

Exercises

I
Charge density and electric flux. Use $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$.
  • 1 A rod $50$ cm long carries $2.0\ \mu\text{C}$ spread uniformly along it. Find its linear charge density.
  • 2 A thin spherical shell of radius $5.0$ cm carries $3.0$ nC. Find its surface charge density.
  • 3 A square of side $20$ cm is in a uniform field of $500$ N/C; its normal makes $30^\circ$ with the field. Find the flux through it.
  • 4 A flat disc of radius $10$ cm is placed in a uniform field of $2.0\times10^3$ N/C. Find the greatest possible flux through it, and the angle between the field and the normal at which the flux is half of that.
  • 5 $\mathbf E=3.0\times10^3\,\hat{\mathbf i}$ N/C. Find the flux through a square of side $10$ cm lying (a) in the $yz$-plane, (b) in the $xy$-plane.
  • 6 $\mathbf E=(4.0\,\hat{\mathbf i}+3.0\,\hat{\mathbf j})\times10^3$ N/C and a flat surface has area vector $\mathbf A=(0.020\,\hat{\mathbf i}+0.050\,\hat{\mathbf j})\ \text{m}^2$. Find the flux through it.
  • 7 Explain why the net flux of a uniform field through any closed surface is zero.
  • 8 Give the SI unit of electric flux in two equivalent forms and its dimensional formula.
1 $\lambda=\dfrac QL=\dfrac{2.0\times10^{-6}}{0.50}=4.0\times10^{-6}$ C/m $=4.0\ \mu\text{C/m}$.
2 $\sigma=\dfrac{Q}{4\pi R^2}=\dfrac{3.0\times10^{-9}}{4\pi(0.050)^2}\approx9.5\times10^{-8}\ \text{C/m}^2$.
3 $\Phi=EA\cos\theta=(500)(0.040)\cos30^\circ\approx17$ N m²/C.
4 Face-on: $\Phi_\text{max}=E\pi R^2=(2.0\times10^3)\pi(0.10)^2\approx63$ N m²/C. Half when $\cos\theta=\tfrac12$: $\theta=60^\circ$.
5 (a) The normal is along $\hat{\mathbf i}$: $\Phi=(3.0\times10^3)(0.010)=30$ N m²/C (sign depends on which way the normal is chosen). (b) The normal is along $\hat{\mathbf k}$, perpendicular to $\mathbf E$: $\Phi=0$.
6 $\Phi=\mathbf E\cdot\mathbf A=(4.0\times10^3)(0.020)+(3.0\times10^3)(0.050)=80+150=230$ N m²/C.
7 Every field line of a uniform field that enters the surface also leaves it (lines do not start or end where there is no charge), so inward (negative) and outward (positive) flux are equal. Equivalently, by Gauss's law, a uniform field needs no charge inside the surface.
8 $\text{N m}^2\,\text{C}^{-1}$, or $\text{V m}$ (since $\text{N/C}=\text{V/m}$). $[\Phi]=[E][A]=[\text{MLT}^{-3}\text{A}^{-1}][\text{L}^2]=[\text{ML}^3\text{T}^{-3}\text{A}^{-1}]$.
II
Gauss's law.
  • 1 A charge of $+5.0\ \mu\text{C}$ is at the centre of a sphere of radius $20$ cm. Find the flux through the sphere. What happens if the radius is doubled?
  • 2 The net flux out of a closed surface is $-1.13\times10^3$ N m²/C. What charge does it enclose?
  • 3 A charge of $5.31$ nC is at the centre of a cube. Find the flux through the whole cube and through one face.
  • 4 A closed surface encloses $+2.0$ nC and $-7.0$ nC; a $+10$ nC charge is just outside. Find the net flux. Is the field on the surface produced only by the two charges inside?
  • 5 For a certain closed surface, $4.0\times10^3$ N m²/C of flux enters and $6.0\times10^3$ N m²/C leaves. Find the enclosed charge.
  • 6 The net flux through a closed surface is zero. Does this mean that the field is zero everywhere on the surface? That there is no charge inside? Give an example.
  • 7 A $1.0\ \mu\text{C}$ charge is at the centre of a sphere. Find the flux through one hemisphere of it (one half of the sphere's surface).
  • 8 A $2.0\ \mu\text{C}$ charge is inside a cube of edge $9.0$ cm. Find the net flux. Does it change if (a) the edge is doubled, (b) the charge is moved off-centre (still inside), (c) the charge is moved outside?
Check: a Gauss's-law calculator for your answers
$q/\varepsilon_0$ (N m²/C)—
one cube face, charge at centre—
$\varepsilon_0\Phi$ (charge from flux)—
1 $\Phi=\dfrac{5.0\times10^{-6}}{8.85\times10^{-12}}\approx5.6\times10^{5}$ N m²/C. Unchanged if the radius is doubled — the enclosed charge is the same.
2 $q=\varepsilon_0\Phi=(8.85\times10^{-12})(-1.13\times10^3)\approx-1.0\times10^{-8}$ C $=-10$ nC.
3 $\Phi=\dfrac{5.31\times10^{-9}}{8.85\times10^{-12}}=600$ N m²/C; each face $100$ N m²/C by symmetry.
4 $q_\text{enc}=-5.0$ nC: $\Phi=\dfrac{-5.0\times10^{-9}}{8.85\times10^{-12}}\approx-565$ N m²/C. No: the field on the surface is the total field, including the large contribution of the $+10$ nC charge; that charge only adds zero to the net flux.
5 Net outward flux $=6.0\times10^3-4.0\times10^3=2.0\times10^3$ N m²/C, so $q=\varepsilon_0\Phi\approx1.8\times10^{-8}$ C $\approx18$ nC.
6 Neither. Zero net flux means zero net enclosed charge. Examples: a dipole inside (field on the surface not zero, charges inside but they sum to zero); a surface with no charge inside but a charge nearby outside (field not zero on the surface).
7 By symmetry each half gets half the total: $\dfrac{q}{2\varepsilon_0}=\dfrac{1.0\times10^{-6}}{2(8.85\times10^{-12})}\approx5.6\times10^{4}$ N m²/C.
8 $\Phi=\dfrac{2.0\times10^{-6}}{8.85\times10^{-12}}\approx2.3\times10^{5}$ N m²/C. (a) No. (b) No — only the flux through individual faces changes. (c) Yes: the net flux becomes zero.
III
Line, sheet and shell. Use $k=9.0\times10^9\ \text{N m}^2\text{C}^{-2}$ and $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$.
  • 1 Find the field $10$ cm from a long straight wire carrying $3.0\ \mu\text{C}$ per metre.
  • 2 At what distance from a long line charge of $5.0$ nC/m is the field $900$ N/C?
  • 3 Find the field near a large sheet with $\sigma=1.77\ \mu\text{C/m}^2$. Does it depend on the distance from the sheet?
  • 4 Two large parallel plates carry $\pm17.7\ \text{nC/m}^2$. Find the field between them and outside them.
  • 5 Two large parallel sheets both carry $+8.85\ \text{nC/m}^2$. Find the field between them and outside them.
  • 6 A thin shell of radius $20$ cm carries $8.0$ nC. Find the field at $10$ cm, just outside the surface, and at $40$ cm from the centre.
  • 7 The field just outside a thin charged shell of radius $5.0$ cm is $3.6\times10^4$ N/C, pointing outwards. Find the charge on the shell and its surface charge density.
  • 8 An electron moves in a circle round a long straight wire carrying $+1.0$ nC/m. Show that its speed does not depend on the radius of the circle, and find it. ($e=1.6\times10^{-19}$ C, $m_e=9.1\times10^{-31}$ kg)
1 $E=\dfrac{2k\lambda}{r}=\dfrac{2(9.0\times10^9)(3.0\times10^{-6})}{0.10}=5.4\times10^{5}$ N/C, radially outwards.
2 $r=\dfrac{2k\lambda}{E}=\dfrac{2(9.0\times10^9)(5.0\times10^{-9})}{900}=0.10$ m.
3 $E=\dfrac{\sigma}{2\varepsilon_0}=\dfrac{1.77\times10^{-6}}{2(8.85\times10^{-12})}=1.0\times10^{5}$ N/C, away from the sheet on both sides — independent of distance (for distances small compared with the sheet).
4 Between: $\dfrac{\sigma}{\varepsilon_0}=\dfrac{17.7\times10^{-9}}{8.85\times10^{-12}}=2.0\times10^{3}$ N/C from the positive to the negative plate. Outside: $0$.
5 Between: the two fields $\sigma/2\varepsilon_0$ are opposite, so $0$. Outside: they add, $\dfrac{\sigma}{\varepsilon_0}=1.0\times10^{3}$ N/C, away from the sheets.
6 At $10$ cm (inside): $0$. Just outside: $\dfrac{(9.0\times10^9)(8.0\times10^{-9})}{(0.20)^2}=1.8\times10^{3}$ N/C. At $40$ cm: a quarter of that, $450$ N/C. Both outwards.
7 $Q=\dfrac{ER^2}{k}=\dfrac{(3.6\times10^4)(0.050)^2}{9.0\times10^9}=1.0\times10^{-8}$ C $=+10$ nC; $\sigma=\dfrac{Q}{4\pi R^2}\approx3.2\times10^{-7}\ \text{C/m}^2$ (check: $\sigma/\varepsilon_0\approx3.6\times10^4$ N/C ✓).
8 The electric force supplies the centripetal force: $\dfrac{m_ev^2}{r}=e\,\dfrac{2k\lambda}{r}$, so $v=\sqrt{\dfrac{2k\lambda e}{m_e}}$ — $r$ cancels. $v=\sqrt{\dfrac{2(9.0\times10^9)(1.0\times10^{-9})(1.6\times10^{-19})}{9.1\times10^{-31}}}\approx1.8\times10^{6}$ m/s.
IV
Thinking with Gauss's law.
  • 1 Explain, using Gauss's law, why a person inside a closed metal car is safe from the field of a lightning strike on the car.
  • 2 Modelling. In fair weather there is a field of about $100$ N/C pointing vertically downwards near the Earth's surface. Treating the Earth as a sphere of radius $6.4\times10^6$ m, estimate the charge on the Earth (with its sign) and the surface charge density.
  • 3 A point charge $q$ sits at the centre of a thin spherical shell of radius $R$ carrying charge $Q$. Find the field at distance $r$ from the centre for $r\lt R$ and $r\gt R$.
  • 4 Two concentric thin shells have radii $10$ cm and $20$ cm and carry $+4.0$ nC and $-6.0$ nC. Find the field at $5$ cm, $15$ cm and $30$ cm from the centre.
  • 5 A long thin wire carrying $+2.0$ nC/m lies along the axis of a long thin cylindrical shell of radius $3.0$ cm carrying $-2.0$ nC per metre of length. Find the field $1.0$ cm from the axis and $5.0$ cm from the axis.
  • 6 In some region the field is $\mathbf E=800\,x\,\hat{\mathbf i}$ N/C (with $x$ in metres). A cube of edge $0.10$ m has faces at $x=0.10$ m and $x=0.20$ m, the others parallel to $\hat{\mathbf i}$. Find the net flux through the cube and the charge inside it.
  • 7 A point charge $q$ is placed a distance $d/2$ directly above the centre of a square of side $d$. Find the flux through the square. (Hint: imagine five more squares.)
  • 8 Sketch $E$ against distance for a thin shell and for an infinite sheet. Show that in both cases the field jumps by $\sigma/\varepsilon_0$ as you cross the charged surface.
1 The charge spreads over the outer metal surface, which is (roughly) a closed shell. Like the shell of Result 1.11, a Gaussian surface drawn inside the car encloses no charge, and the symmetry argument — exact for a sphere, very nearly true for a closed metal body (Section 2.3) — gives zero field inside. The current of the strike also flows round the outside to the ground.
2 The field points inwards, so the Earth is negative. Gauss's law over a sphere just above the ground: $E(4\pi R^2)=|Q|/\varepsilon_0$, i.e. $|Q|=\dfrac{ER^2}{k}=\dfrac{(100)(6.4\times10^6)^2}{9.0\times10^9}\approx4.6\times10^{5}$ C, so $Q\approx-4.6\times10^5$ C. $\sigma=-\varepsilon_0E\approx-8.9\times10^{-10}\ \text{C/m}^2$.
3 Inside, a concentric Gaussian sphere encloses only $q$: $E=\dfrac{kq}{r^2}$. Outside it encloses $q+Q$: $E=\dfrac{k(q+Q)}{r^2}$. Both radial.
4 $5$ cm: encloses nothing, $E=0$. $15$ cm: encloses $+4.0$ nC, $E=\dfrac{(9.0\times10^9)(4.0\times10^{-9})}{(0.15)^2}=1.6\times10^{3}$ N/C outwards. $30$ cm: encloses $-2.0$ nC, $E=\dfrac{(9.0\times10^9)(2.0\times10^{-9})}{(0.30)^2}=200$ N/C, inwards.
5 At $1.0$ cm a coaxial cylinder encloses only the wire: $E=\dfrac{2k\lambda}{r}=\dfrac{2(9.0\times10^9)(2.0\times10^{-9})}{0.010}=3.6\times10^{3}$ N/C outwards. At $5.0$ cm it encloses $\lambda-\lambda=0$: $E=0$. (This is how a coaxial cable keeps its field inside.)
6 Flux only through the two faces perpendicular to $\hat{\mathbf i}$: $\Phi=E(0.20)A-E(0.10)A=800(0.20-0.10)(0.010)=0.80$ N m²/C. $q=\varepsilon_0\Phi\approx7.1\times10^{-12}$ C.
7 The square is one face of a cube of side $d$ with $q$ at its centre. By symmetry each of the six faces gets an equal share: $\Phi=\dfrac{q}{6\varepsilon_0}$.
8 Shell: $E=0$ inside, $\sigma/\varepsilon_0$ just outside (Result 1.11), then $\propto1/r^2$ — a jump of $\sigma/\varepsilon_0$. Sheet: $E_x=-\sigma/2\varepsilon_0$ on one side and $+\sigma/2\varepsilon_0$ on the other — also a jump of $\sigma/\varepsilon_0$. (A pillbox straddling any small patch of charged surface gives the same jump.)
🎯 Exam corner — flux and Gauss's law BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks State Gauss's law in electrostatics. Use it to find the electric field due to an infinite plane sheet of charge with uniform surface charge density $\sigma$. The field $5.0$ cm from such a sheet is $2.0\times10^4$ N/C; find $\sigma$ and the field $10$ cm from the sheet.

Law The net flux out of any closed surface equals $q_\text{enc}/\varepsilon_0$: $\oint\mathbf E\cdot d\mathbf A=q_\text{enc}/\varepsilon_0$.
Setup By symmetry $\mathbf E$ is perpendicular to the sheet, away from it on both sides, with equal magnitude at equal distances. Gaussian surface: a cylinder of end area $A$ piercing the sheet at right angles (diagram).
Equations Curved surface: $\mathbf E\perp d\mathbf A$, no flux. Two ends: $EA$ each. Enclosed charge $\sigma A$. So $2EA=\sigma A/\varepsilon_0$, giving $$E=\frac{\sigma}{2\varepsilon_0}.$$
Numbers $\sigma=2\varepsilon_0E=2(8.85\times10^{-12})(2.0\times10^4)\approx3.5\times10^{-7}\ \text{C/m}^2$. At $10$ cm the field is the same, $2.0\times10^4$ N/C: it does not depend on distance. Exam tip: marks are given for stating why the curved surface contributes no flux.

E2. Board · Case study A student puts a point charge $q=+1.77\ \mu\text{C}$ at the centre of an imaginary cube of edge $10$ cm. (i) Find the total flux out of the cube and the flux through one face. (2 marks) (ii) The charge is moved $2$ cm towards one face, still inside. Which of the answers to (i) change? (1 mark) (iii) A charge $-1.77\ \mu\text{C}$ is now placed just outside the cube. What is the total flux out of the cube? (1 mark)

(i) $\Phi=\dfrac{q}{\varepsilon_0}=\dfrac{1.77\times10^{-6}}{8.85\times10^{-12}}=2.0\times10^{5}$ N m²/C; each face $\dfrac{\Phi}{6}\approx3.3\times10^{4}$ N m²/C.
(ii) The total flux is unchanged (same enclosed charge). The flux through individual faces changes — the near face gets more, the far face less — because the symmetry is lost.
(iii) The outside charge contributes nothing to the net flux: still $2.0\times10^5$ N m²/C. Exam tip: case studies often test exactly this distinction between total flux and flux through one face.

E3. JEE Main An infinitely long thin wire with uniform linear charge density produces a field of $9.0\times10^4$ N/C at a distance of $2.0$ cm. Its linear charge density is (A) $0.10\ \mu\text{C/m}$ (B) $0.20\ \mu\text{C/m}$ (C) $1.0\ \mu\text{C/m}$ (D) $0.050\ \mu\text{C/m}$.

Concept $E=\dfrac{2k\lambda}{r}$, so $\lambda=\dfrac{Er}{2k}$.
Numbers $\lambda=\dfrac{(9.0\times10^4)(0.020)}{2(9.0\times10^9)}=1.0\times10^{-7}$ C/m.
Answer (A) $0.10\ \mu\text{C/m}$. Exam tip: (B) is what you get by using $k\lambda/r$ — dropping the factor $2$ in $2k\lambda/r$ is a classic slip.

E4. NEET A thin spherical shell of radius $R$ carries a charge $Q$ uniformly. The magnitudes of the field at distances $R/2$ and $2R$ from its centre are (A) $0$ and $\dfrac{kQ}{4R^2}$ (B) $\dfrac{4kQ}{R^2}$ and $\dfrac{kQ}{4R^2}$ (C) $\dfrac{kQ}{2R^2}$ and $\dfrac{kQ}{4R^2}$ (D) $0$ and $0$.

Concept Result 1.11: zero inside a thin shell; $kQ/r^2$ outside.
Values At $R/2$: inside, $E=0$. At $2R$: $E=\dfrac{kQ}{(2R)^2}=\dfrac{kQ}{4R^2}$.
Answer (A). Exam tip: (B) treats the shell as a point charge everywhere — true only outside.

E5. JEE Advanced (One or more correct.) Two large parallel sheets $P$ (left) and $S$ (right) carry uniform surface charge densities $+2\sigma$ and $-\sigma$ ($\sigma\gt0$). Which statements are correct? (A) Between the sheets the field is $\dfrac{3\sigma}{2\varepsilon_0}$, from $P$ towards $S$. (B) The field has the same magnitude, $\dfrac{\sigma}{2\varepsilon_0}$, to the left of $P$ and to the right of $S$. (C) The field to the right of $S$ is zero. (D) To the left of $P$ the field points away from the sheets.

Concept Each sheet alone gives a uniform field $|\sigma_i|/2\varepsilon_0$, away from a positive sheet and towards a negative one; superpose region by region. Take $+x$ to the right.
Regions Left of $P$: $P$ gives $-\dfrac{\sigma}{\varepsilon_0}$, $S$ gives $+\dfrac{\sigma}{2\varepsilon_0}$; net $-\dfrac{\sigma}{2\varepsilon_0}$ (to the left). Between: $+\dfrac{\sigma}{\varepsilon_0}+\dfrac{\sigma}{2\varepsilon_0}=+\dfrac{3\sigma}{2\varepsilon_0}$. Right of $S$: $+\dfrac{\sigma}{\varepsilon_0}-\dfrac{\sigma}{2\varepsilon_0}=+\dfrac{\sigma}{2\varepsilon_0}$.
Answer (A) true, (B) true, (C) false, (D) true: (A), (B), (D). Exam tip: outside, the pair behaves like one sheet of net density $+\sigma$, giving $\sigma/2\varepsilon_0$ pointing away on both sides — a quick check.

E6. JEE Main (Numerical value.) A point charge of $5.31$ nC is placed at a height $a/2$ directly above the centre of a square of side $a$. The electric flux through the square, in N m²/C, is ______. ($\varepsilon_0=8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$)

Idea The square is one face of a cube of side $a$ centred on the charge; by symmetry it receives one-sixth of the total flux.
Numbers $\Phi=\dfrac{q}{6\varepsilon_0}=\dfrac{5.31\times10^{-9}}{6(8.85\times10^{-12})}=\dfrac{5.31\times10^{-9}}{5.31\times10^{-11}}=100$.
Answer 100 N m²/C. Exam tip: the side $a$ never enters — if a problem gives a length you do not need, look for a symmetric closed surface.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. The fair-weather electric field of the atmosphere gets weaker with height. What does that tell us, through Gauss's law, about where the charge of the "global electric circuit" is?

Why it matters. Thunderstorms around the world act like a giant battery, keeping the ground negative and the upper atmosphere positive; in fair-weather regions a tiny current leaks back down through the air. Atmospheric physicists measure the field with rotating "field mills" at observatories from Scotland to Antarctica and on aircraft and balloons; pollution, cosmic rays and climate all change it. Famously, the global field rises and falls with a daily rhythm that follows universal time, not local time, first mapped from measurements on the research ship Carnegie in the 1920s.

What to do.

  1. Sample data — the downward fair-weather field at several heights: ground $120$ N/C, $1$ km $40$ N/C, $5$ km $12$ N/C, $10$ km $4$ N/C.
  2. Over such small heights the Earth is flat: use a vertical pillbox of area $1\ \text{m}^2$ between two heights. Show that Gauss's law gives the charge per square metre of ground between them as $\varepsilon_0(E_\text{lower}-E_\text{upper})$, and find it for each layer.
  3. Find the surface charge density on the ground, $\sigma=-\varepsilon_0E_\text{ground}$, and the total charge of the Earth.
  4. Compare the charge in the air below $10$ km with the charge on the ground. What would a satellite far above the atmosphere measure?

Tools: Calculator or spreadsheet.

Hand in: A table of layer charges and average charge densities (in C/m³ and elementary charges per cm³), the Earth's charge, and a paragraph answering the satellite question.

Going further. Air conducts slightly because cosmic rays ionise it. If the fair-weather current density is about $2\times10^{-12}\ \text{A/m}^2$, how long would it take to discharge the ground if the thunderstorms stopped?

Instructor note: layer charges per m² are $\varepsilon_0\times80\approx7.1\times10^{-10}$ C (0–1 km), $\varepsilon_0\times28\approx2.5\times10^{-10}$ C (1–5 km), $\varepsilon_0\times8\approx7.1\times10^{-11}$ C (5–10 km), all positive; total $\approx1.03\times10^{-9}$ C/m². Average density in the lowest km $\approx7.1\times10^{-13}$ C/m³, about $4$ elementary charges per cm³. Ground: $\sigma\approx-1.06\times10^{-9}$ C/m², Earth's charge $\approx-5.5\times10^5$ C. The air below 10 km almost cancels the ground's charge (the residual field at 10 km is only $4$ N/C), so from far away the Earth plus atmosphere is nearly neutral. Discharge time $\sim|\sigma|/j\approx530$ s, about nine minutes — storms must recharge it continuously.

🧪 Self-check quiz

0/5 correct

Q1. A closed surface encloses charges $+q$ and $-q$ and nothing else. The net flux out of it is

Q2. The SI unit of electric flux is

Q3. The field of an infinitely long uniform line charge varies with distance $r$ from it as

Q4. Inside a uniformly charged thin spherical shell, the electric field is

Q5. Two large parallel plates carry surface charge densities $+\sigma$ and $-\sigma$. The field between them is

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