Fields of Continuous Charge

A continuous charge is simply charge spread out smoothly over a wire, a ring or a plate, instead of sitting at a few separate points. "Continuous" because there are no gaps: the charge is smeared so finely that we describe it by how much there is per meter or per square meter, like paint rather than dots. That's what real objects look like, from power lines to capacitor plates, and all we're really doing is adding up a lot of tiny point charges. The formula $\mathbf{E} = \int \frac{k\,dq}{r^2}\,\hat{\mathbf{r}}$ looks scary, but it's just the point-charge field with a few small changes: $dq$ (one tiny crumb of the charge), $\frac{k}{r^2}$ (how strong that crumb's field is from where you're standing), $\hat{\mathbf{r}}$ (which way it points) and $\int$ (add up every crumb). That's it. The integrals get a little long, but each one is the same simple idea, so bear with them.

    **The recipe.** So far every charge has been a point. But a charged wire, ring or plate is really a crowd of point charges packed too tightly to count. We already know what to do with a crowd of point charges: superposition, adding up their fields one at a time. The only change is that now there are infinitely many tiny pieces, and adding up infinitely many tiny pieces is exactly what an integral does. Chop the object into pieces of charge $dq$, give each the point-charge field $k\,dq/r^2$ and integrate:

    1. Choose coordinates and write $dq$ in terms of the variable you'll integrate: $dq = \lambda\,dx$ along a wire, $dq = \lambda R\,d\varphi$ around a ring, $dq = \sigma\cdot2\pi s\,ds$ for a thin ring-shaped strip of a disk.
    2. Write the vector from $dq$ to the field point and its length $r$.
    3. Use symmetry to see which components cancel.
    4. Integrate each surviving component, never the magnitude, since every piece's field points a different way.
    5. Check the answer far away, where any finite object must look like a point charge, $kQ/r^2$.

    **A straight segment.** Put a segment of length $L$ with uniform density $\lambda$ on the $x$-axis from $-L/2$ to $L/2$, and the field point a distance $d$ away on its perpendicular bisector. The pieces at $x$ and $-x$ cancel each other's components along the segment. The piece at $x$ is $r = \sqrt{x^2 + d^2}$ away, and the part of its field perpendicular to the segment is $\frac{k\lambda\,dx}{r^2}\cdot\frac dr$, so
    $$E = \int_{-L/2}^{L/2}\frac{k\lambda d\,dx}{(x^2 + d^2)^{3/2}}.$$
    The [trig substitution](/learn/calculus-2/lessons/trig-substitution) $x = d\tan\theta$ gives $dx = d\sec^2\theta\,d\theta$ and $(x^2 + d^2)^{3/2} = d^3\sec^3\theta$, so the integrand collapses to $\frac{k\lambda}{d}\cos\theta\,d\theta$ and
    $$E = \frac{k\lambda}{d}\Big[\sin\theta\Big]_{-\theta_0}^{\theta_0} = \frac{2k\lambda}{d}\cdot\frac{L/2}{\sqrt{d^2 + (L/2)^2}}.$$

Now check it. Far away ($d \gg L$) the fraction is about $\frac{L}{2d}$, so $E \approx \frac{k\lambda L}{d^2} = \frac{kQ}{d^2}$: from far off the rod looks like a point, just as it should. For an endless wire ($L \to \infty$) the fraction tends to 1 and $E = 2k\lambda/d$. Notice that fades only like $1/d$, slower than a point charge. That makes sense, because however far you back away, there's always more wire just out of reach.

    **A ring.** A ring of radius $R$ carries total charge $Q$, and the field point is on its axis, $z$ from the center. Every piece of the ring is the same distance $r = \sqrt{z^2 + R^2}$ away, and the piece directly across the ring cancels its sideways component. The axial component, $\frac{k\,dq}{r^2}\cdot\frac zr$, is the same for every piece, so the integral just adds up the charge:
    $$E_z = \frac{kz}{(z^2 + R^2)^{3/2}}\int dq = \frac{kQz}{(z^2 + R^2)^{3/2}}.$$
    Two quick checks. It's zero at the center, where every piece is cancelled by the one across from it, and it's about $kQ/z^2$ far away, where the ring looks like a point.

    **A disk** of radius $R$ and surface charge density $\sigma$ is a nest of rings. The ring of radius $s$ and width $ds$ holds $dq = \sigma\,2\pi s\,ds$, so, substituting $u = z^2 + s^2$,
    $$E_z = \int_0^R\frac{kz\,\sigma\,2\pi s\,ds}{(z^2 + s^2)^{3/2}} = \pi k\sigma z\int_{z^2}^{z^2 + R^2}u^{-3/2}\,du = 2\pi k\sigma\left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right).$$
    Now let the disk grow ($R \to \infty$): the fraction goes to 0 and $E = 2\pi k\sigma = \frac{\sigma}{2\varepsilon_0}$. That's surprising at first. An infinite sheet's field doesn't depend on distance at all. Here's one way to see why: as you back away, each piece gets weaker, but more and more of the sheet points nearly straight at you instead of sideways, and the two effects exactly balance. Close to any large plate ($z \ll R$) you're in that regime. Two oppositely charged plates make $\sigma/\varepsilon_0$ between them and nothing outside: a uniform field, the workhorse of the [next lesson](/learn/electromagnetism/lessons/charges-in-fields).

    **A common mistake.** It's tempting to integrate $k\,dq/r^2$ straight, adding up the strengths, because that's the formula for each piece. That only works when every piece's field points the same way. Usually they don't, so resolve each piece's field into components first and integrate those. Along the same lines, $r$ and the angle change from piece to piece along a wire; on a ring's axis they happen not to, which is what made the ring so easy. Also watch the bookkeeping: a thin ring-shaped strip of a disk has area $2\pi s\,ds$, not $ds$, and the formulas want $\lambda = Q/L$ or $\sigma = Q/A$, not $Q$ itself.

    The same integrals give the masses and centers of mass of wires and plates, as a [double integral](/learn/calculus-3/lessons/double-integrals) or a [surface integral](/learn/calculus-3/lessons/surface-integrals-scalar). And [Gauss's Law](/learn/electromagnetism/lessons/gauss-law) will produce the infinite line and the infinite sheet in a single line each, with no integration at all.

Engineers model power lines as line charges to find the fields that people and equipment beneath them experience, and treat wide films, belts and plates as charged sheets to predict static cling and sparking. Inkjet deflection plates, capacitors and the electrostatic chucks that hold silicon wafers flat during chip fabrication all rely on the sheet result: near a large plate, the field doesn't depend on distance. Simulation software for sensors and high-voltage insulators does this lesson's integral numerically, chopping surfaces into millions of tiny charged patches.