Continuous Charge Distributions

Last lesson you added up the fields of a handful of point charges. Three charges, five, maybe ten: write each $k q/r^2$, resolve into components, add. Tedious, but nothing mysterious.

Now the charge is smeared. A plastic rod rubbed with wool. A ring of wire. The drum of a photocopier. There is no "the charge" sitting at a point, there are $10^{12}$ of them spread over something, and the sum you want has no last term. The trick is that nothing new happens physically. Every tiny piece of that rod is still a point charge, still obeys Coulomb's law, and the fields still add. The sum just becomes an integral.

Chop it into point charges

Take a chunk of the object so small that it may as well be a point. Call its charge $dq$. It sits a distance $r$ from the place you care about, and it makes a field there of

$d\vec E = \frac{k\,dq}{r^2}\,\hat r$

where $k = 8.99\times10^9$ N·m²/C² and $\hat r$ is the unit vector pointing from the chunk towards your field point. Add up every chunk:

$\vec E = \int d\vec E$

That is the whole physics. It is one line and it is already finished. What remains is bookkeeping, but the bookkeeping is where people lose the problem, so it is worth having a fixed routine.

How much charge is in a chunk depends on how the charge is spread. On a wire you quote charge per unit length, $\lambda$, in C/m, so $dq = \lambda\,dx$. On a surface you quote charge per unit area, $\sigma$, in C/m², so $dq = \sigma\,dA$. Through a solid you quote charge per unit volume, $\rho$, in C/m³, so $dq = \rho\,dV$. Those three letters are the whole vocabulary.

The three decisions

One: choose the element. Pick a single variable that sweeps the object exactly once, and write $dq$ in terms of it. For a rod, the position $x$ along it. For an arc, the angle $\phi$. For a disk, the radius $s$ of a thin ring inside it, and that last choice matters, because if you already know a ring's field, a disk is just a stack of rings and you never have to do a double integral.

Two: kill a component before you integrate. $d\vec E$ is a vector, and you cannot integrate a vector by integrating its length. Look for symmetry first. If every element has a partner whose sideways contribution points the opposite way, those sideways parts cancel in pairs, and you only need the surviving component: $dE_\parallel = dE\cos\theta$. Now you are integrating a scalar, which you know how to do.

Three: put everything in the one variable. Both $r$ and $\cos\theta$ have to be rewritten in terms of your integration variable, with limits that cover the object once and only once. If an $r$ or a $\theta$ is still floating around when you start integrating, the answer will be wrong.

The ring, and why it is the one to memorise

Put a ring of radius $R$ with total charge $Q$ spread uniformly around it in the $xy$-plane, and stand at a point $P$ on the axis, a distance $z$ from the centre.

Every element of the ring is the same distance from $P$: $r = \sqrt{z^2 + R^2}$. That is the reason this case is easy. Each element makes a field of magnitude $k\,dq/(z^2+R^2)$ pointing from the element towards $P$, tilted off the axis. But the element directly opposite makes an equally tilted field the other way, so the off-axis parts cancel in pairs. What survives is the axial part, $dE\cos\theta$, with

$\cos\theta = \frac{z}{\sqrt{z^2+R^2}}$

Since $r$ and $\cos\theta$ are the same for every element, both come out of the integral and all that is left is $\int dq = Q$:

$E_z = \frac{kQz}{(z^2+R^2)^{3/2}}$

Read it. At $z = 0$ it gives zero, which it must: at the centre every element is cancelled by the one across from it. For $z \gg R$ the $R^2$ is negligible and it becomes $kQ/z^2$: a point charge, as it has to be from far away. Between those two the field rises and falls, so somewhere it peaks. Setting $dE_z/dz = 0$ gives $(z^2+R^2)^{3/2} = 3z^2(z^2+R^2)^{1/2}$, so $z^2 + R^2 = 3z^2$ and

$z_{\text{peak}} = \frac{R}{\sqrt 2}$

Example 1. A ring of radius $0.10$ m carries $5.0$ nC. Find $E$ on the axis at $z = 0.20$ m.

$z^2 + R^2 = 0.040 + 0.010 = 0.050$ m². Then $(0.050)^{3/2} = 0.050 \times \sqrt{0.050} = 0.050 \times 0.2236 = 0.01118$. The numerator is $kQz = (8.99\times10^9)(5.0\times10^{-9})(0.20) = 8.99$. So $E = 8.99/0.01118 = 804$ N/C, pointing away from the ring along the axis.

There is one more thing hiding in the ring formula. Very close to the centre, $z \ll R$, the denominator is just $R^3$, so $E_z \approx kQz/R^3$: a field that grows linearly with displacement. A charge $-q$ of mass $m$ on the axis then feels a restoring force $-kQqz/R^3$, which is Hooke's law with $k_{\text{eff}} = kQq/R^3$. It oscillates, with $\omega = \sqrt{kQq/(mR^3)}$. That is the whole solution to any "bead through a ring" problem you meet.

Straight rods

A rod has no such luck: different elements sit at different distances. Put the rod along the $y$-axis from $-L/2$ to $+L/2$ with uniform $\lambda = Q/L$, and stand at $P$ on the perpendicular bisector, a distance $y$ away.

The element at height $x$ carries $dq = \lambda\,dx$ and sits a distance $r = \sqrt{x^2+y^2}$ from $P$. Its field at $P$ has a component along the bisector and a component parallel to the rod. The element at $-x$ gives the same thing with the parallel component flipped, so those cancel in pairs and only the bisector component survives, with $\cos\theta = y/\sqrt{x^2+y^2}$. So

$E = \int_{-L/2}^{L/2} \frac{k\lambda\,dx}{x^2+y^2}\cdot\frac{y}{\sqrt{x^2+y^2}} = k\lambda y\int_{-L/2}^{L/2}\frac{dx}{(x^2+y^2)^{3/2}}$

That integral is a standard one, and it is worth knowing where it comes from: substitute $x = y\tan\theta$, so $dx = y\sec^2\theta\,d\theta$ and $(x^2+y^2)^{3/2} = y^3\sec^3\theta$, and the integrand collapses to $\cos\theta\,d\theta/y^2$. Integrating gives $\sin\theta/y^2$, and $\sin\theta = x/\sqrt{x^2+y^2}$. So

$E = \frac{k\lambda}{y}\left[\frac{x}{\sqrt{x^2+y^2}}\right]_{-L/2}^{L/2} = \frac{2k\lambda}{y}\cdot\frac{L/2}{\sqrt{y^2+L^2/4}} = \frac{kQ}{y\sqrt{y^2+L^2/4}}$

Two limits fall straight out. If $y \gg L$ the square root is just $y$ and you get $kQ/y^2$: a point charge again. If $y \ll L$ the square root is $L/2$ and the $L$ cancels, leaving

$E_{\text{infinite wire}} = \frac{2k\lambda}{y}$

An infinite line falls off like $1/y$, not $1/y^2$. That is worth carrying around; you will derive it again in one line with Gauss's law.

The same integral, with different limits, does other rods. For a half-infinite rod running from $x=0$ to $x=\infty$, the limits become $0$ to $\infty$, which in the $\theta$ form runs from $0$ to $90°$. The parallel component no longer cancels, so you need both: the perpendicular part is $k\lambda/y$ and the parallel part is also $k\lambda/y$, giving a field of magnitude $\sqrt2\,k\lambda/y$ tilted at 45°. And for a point on the rod's own axis there is no vector work at all (every element pushes the same way) so $E = \int_a^b k\lambda\,dx/x^2 = k\lambda(1/a - 1/b)$.

Example 2. A rod of length $0.40$ m carries $15$ nC uniformly. Find $E$ at a point on the perpendicular bisector $0.10$ m from the rod.

Use $E = kQ/\big(y\sqrt{y^2+L^2/4}\big)$. Here $L/2 = 0.20$ m, so $y^2 + L^2/4 = 0.010 + 0.040 = 0.050$ m² and $\sqrt{0.050} = 0.2236$ m. The numerator is $kQ = (8.99\times10^9)(15\times10^{-9}) = 134.85$. The denominator is $(0.10)(0.2236) = 0.02236$. So $E = 134.85/0.02236 = 6.03\times10^3$ N/C.

Compare that with the infinite-wire estimate: $\lambda = 15/0.40 = 37.5$ nC/m, so $2k\lambda/y = 2(8.99\times10^9)(3.75\times10^{-8})/0.10 = 6.74\times10^3$ N/C. Close, and too big: the real rod is missing the far-away charge the infinite wire has. Checking against a limit like that catches sign and factor errors fast.

Disks, washers, and the sheet

Now use the ring result instead of starting over. A disk of radius $R$ with uniform $\sigma$ is a stack of rings. The ring of radius $s$ and thickness $ds$ has area $2\pi s\,ds$, so $dq = \sigma\,2\pi s\,ds$, and its contribution on the axis is the ring formula with $R$ replaced by $s$:

$E_z = \int_0^R \frac{k z\,(\sigma 2\pi s\,ds)}{(z^2+s^2)^{3/2}} = \pi k\sigma z\int_0^R \frac{2s\,ds}{(z^2+s^2)^{3/2}}$

The substitution $u = z^2+s^2$ makes $2s\,ds = du$, and $\int u^{-3/2}du = -2u^{-1/2}$, so

$E_z = 2\pi k\sigma\left(1 - \frac{z}{\sqrt{z^2+R^2}}\right) = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{z}{\sqrt{z^2+R^2}}\right)$

using $2\pi k = 1/(2\varepsilon_0)$ with $\varepsilon_0 = 8.85\times10^{-12}$ F/m. Let $R\to\infty$, or equivalently stand very close, and the bracket goes to 1:

$E_{\text{sheet}} = \frac{\sigma}{2\varepsilon_0}$

which does not depend on distance at all. Two parallel plates with $+\sigma$ and $-\sigma$ each contribute this, in the same direction between them and in opposite directions outside, so the gap field is $\sigma/\varepsilon_0$ and the outside field is zero. The same $\sigma/\varepsilon_0$ shows up just outside a charged conductor, which is why a photocopier drum is worked with that formula and not the plastic-sheet one.

A washer, or annulus, is a disk with a smaller disk removed, so its field is the difference of two disk fields with the same $z$:

$E = \frac{\sigma}{2\varepsilon_0}\left(\frac{z}{\sqrt{z^2+a^2}} - \frac{z}{\sqrt{z^2+b^2}}\right)$

for inner radius $a$ and outer radius $b$. Subtracting one whole object from another is often faster than any integral.

Example 3. A disk of radius $R = 8.0$ cm carries a surface density that is not uniform: $\sigma(r) = \sigma_0 r/R$ with $\sigma_0 = 5.0\times10^{-7}$ C/m². Find $E$ on the axis at $z = 6.0$ cm.

You cannot use the uniform-disk answer, but the ring decomposition still works, because $\sigma$ is constant around any one ring. The ring at radius $s$ carries $dq = \sigma(s)\,2\pi s\,ds = 2\pi\sigma_0 s^2\,ds/R$, so

$E_z = \frac{2\pi k\sigma_0 z}{R}\int_0^R\frac{s^2\,ds}{(z^2+s^2)^{3/2}}$

That integral is the other standard one: $\int s^2(z^2+s^2)^{-3/2}ds = \ln\!\big(s+\sqrt{s^2+z^2}\big) - s/\sqrt{s^2+z^2}$. With $R = 0.080$ m and $z = 0.060$ m, $\sqrt{R^2+z^2} = 0.100$ m. The log term between $0$ and $R$ is $\ln\big((0.080+0.100)/0.060\big) = \ln(3.000) = 1.0986$, and the subtracted term is $0.080/0.100 = 0.800$. So the integral is $1.0986 - 0.800 = 0.2986$.

The prefactor: $2\pi k = 5.649\times10^{10}$, times $\sigma_0 = 5.0\times10^{-7}$ gives $2.824\times10^4$, times $z/R = 0.060/0.080 = 0.750$ gives $2.118\times10^4$.

$E_z = (2.118\times10^4)(0.2986) = 6.3\times10^3\ \text{N/C}$

Arcs, and adding pieces as vectors

For an arc, use the angle. A wire bent into a circular arc of radius $R$ with uniform $\lambda$, and the field point at the centre: every element is the same distance $R$ away, so each contributes $k\lambda R\,d\phi/R^2 = k\lambda\,d\phi/R$, pointing radially inward or outward. Only the component along the arc's symmetry axis survives, so integrate $\cos\phi$ over the arc. For a semicircle, $\phi$ runs from $-90°$ to $+90°$ and $\int\cos\phi\,d\phi = 2$:

$E_{\text{semicircle}} = \frac{2k\lambda}{R}$

A quarter circle would give $\sqrt2\,k\lambda/R$ by the same integral, and a full circle gives zero, as it must.

Once you have pieces, combine them as vectors, not as numbers. A ring with $+q$ on one half and $-q$ on the other is two semicircles whose fields at the centre point the same way (away from the positive half, towards the negative half), so they add to $4k\lambda/R$. A shape made of an arc plus a straight bit needs a proper vector sum, because the two pieces usually pull in perpendicular directions.

Example 4. A wire with $\lambda = 3.0\times10^{-8}$ C/m is a semicircle of radius $0.10$ m centred on $O$, plus a straight segment of length $0.10$ m running outward from one end of the semicircle along the diameter line. Find $E$ at $O$.

Do the pieces separately. The semicircle gives $2k\lambda/R = 2(8.99\times10^9)(3.0\times10^{-8})/0.10 = 5394$ N/C, directed along the semicircle's symmetry axis: that is, perpendicular to the diameter.

The straight segment lies along the diameter line, with $O$ on its extension. Its near end is at $0.10$ m from $O$ and its far end at $0.20$ m, so it is the end-on case: $E = k\lambda(1/a - 1/b) = (8.99\times10^9)(3.0\times10^{-8}) (1/0.10 - 1/0.20) = 269.7 \times 5.00 = 1349$ N/C, directed along the diameter.

The two are perpendicular, so add in quadrature:

$E = \sqrt{5394^2 + 1349^2} = \sqrt{2.909\times10^7 + 1.82\times10^6} = 5.56\times10^3\ \text{N/C}$

Adding $5394 + 1349$ would have been wrong by 25 per cent.

Add and subtract whole objects

Superposition works on objects, not only on point charges, and that turns some frightening problems into subtraction.

A hemispherical shell is a stack of rings, and doing that integral gives $\sigma/4\varepsilon_0$ at the centre of its rim: exactly half the full sheet value, which is a reasonable thing for half a surface to give.

A uniformly charged solid ball with a spherical cavity hollowed out of it is a full ball of density $+\rho$ plus a small ball of density $-\rho$ filling the cavity. Inside a uniform ball the field at displacement $\vec r$ from its centre is $\rho\vec r/(3\varepsilon_0)$: you will prove that next lesson. Adding the two contributions at a point inside the cavity, the $\vec r$ terms cancel and only the displacement $\vec d$ between the two centres survives:

$\vec E = \frac{\rho\vec d}{3\varepsilon_0}$

The field inside the cavity is uniform. That is a genuinely surprising result and it comes out of two lines of subtraction.

Recognising which tool the problem wants

Read the geometry first, not the numbers.

If every piece of charge is the same distance from the field point (a ring on its axis, an arc at its centre, a shell at a point outside) the integral is easy and often you only need $\int dq = Q$ with one $\cos\theta$.

If the problem says "on the axis" and the object is flat and round, build it from rings. If it says the density varies with radius, build it from rings anyway; only the $dq$ changes.

If the object is a straight segment, decide first whether the field point is on the perpendicular bisector (one component cancels), on the rod's own axis (no vector work at all), or somewhere general (you need both components). If the words "very long" or "infinite" appear, use $2k\lambda/y$ and skip the integral. If "large sheet" appears, use $\sigma/2\varepsilon_0$, or $\sigma/\varepsilon_0$ if it is a conductor surface or the gap between opposite plates.

If the object is a complete shape with a bite out of it (a washer, a ball with a cavity, a ring missing an arc) subtract rather than integrate.

And if the object is a sphere, or a very long cylinder, or an infinite sheet, and the question asks for the field at a general point rather than just on an axis, stop setting up an integral. That symmetry is exactly what Gauss's law exists to exploit, and it gets those answers in a line.