Electric Fields of Charge Distributions

Coulomb's law gives the field of a point charge (topic 8.3), but a charged rod or ring is not a point: its charge sits at many different distances and directions from the spot where you want the field. The fix is to cut the object into pieces small enough to be point charges, write the field of one piece, and add all of them as vectors. Because there are infinitely many pieces, the sum is an integral.

Building a field from point-charge pieces

Charge spread along a thin object is described by its linear charge density $\lambda$, the charge per length in $\text{C/m}$. A piece of length $d\ell$ carries $dq = \lambda\,d\ell$. (A surface uses $\sigma$ in $\text{C/m}^2$ and a volume uses $\rho$ in $\text{C/m}^3$, with $dq = \sigma\,dA$ or $dq = \rho\,dV$.) For a uniform rod of charge $Q$ and length $L$, $\lambda = \frac{Q}{L}$.

The integral of a vector is not the integral of magnitudes: two pieces whose fields point in different directions partly cancel, and only the components keep track of that.

The cleanest case has no cancellation at all. Take a uniform rod of charge $Q > 0$ along the $x$-axis from $x = 0$ to $x = L$, and find the field at a point on the same line, a distance $d$ past the right end, at $x = L + d$. Every piece pushes a positive test charge in the $+x$ direction, so the components all add. The piece at position $x$ is a distance $L + d - x$ from $P$:

directed in $+x$. Check the limit: when $d \gg L$, $d(L + d) \approx d^2$ and the result becomes the point-charge field $\frac{Q}{4\pi\varepsilon_0 d^2}$. Seen from far away, any finite object looks like a point, and a correct answer must reproduce that.

Letting symmetry cancel components

Before integrating, ask what the arrangement looks like after a reflection or rotation that leaves the charge unchanged. The field must be unchanged by that same operation, and that often forces a component to be zero. In practice you pair each piece with a mirror-image partner: their fields have equal magnitude, and the components perpendicular to the mirror line cancel.

Take a uniform rod of length $L$ and charge $Q$ along the $x$-axis from $-\frac{L}{2}$ to $\frac{L}{2}$, and a point $P$ a distance $y$ from its center on the perpendicular bisector.

The piece at $x$ is a distance $r = \sqrt{x^2 + y^2}$ from $P$, and its field makes angle $\varphi$ with the $y$-axis, where $\cos\varphi = \frac{y}{r}$. The piece at $-x$ has the same $r$ and the opposite horizontal component, so $E_x = 0$ and only the $y$-components survive:

The substitution $x = y\tan\varphi$ makes this easy: $dx = y\sec^2\varphi\,d\varphi$ and $(x^2 + y^2)^{3/2} = y^3\sec^3\varphi$, so the integrand becomes $\frac{\cos\varphi\,d\varphi}{y^2}$ and the integral is $\frac{\sin\varphi}{y^2}$ evaluated between the angles to the two ends. With $\sin\varphi = \frac{x}{\sqrt{x^2 + y^2}}$,

directed away from the rod along the bisector for $Q > 0$.

Two limits tell you what this formula means.

• Far away ($y \gg L$): $\sqrt{y^2 + L^2/4} \approx y$, so $E \approx \frac{Q}{4\pi\varepsilon_0 y^2}$, a point charge.
• Very close, or an infinitely long line ($y \ll L$, or $L \to \infty$ with $\lambda$ fixed): $\frac{L}{\sqrt{y^2 + L^2/4}} \to 2$, so

The field of an infinite line falls off as $\frac{1}{y}$, not $\frac{1}{y^2}$: moving away from a line, you get farther from every piece, but more of the line's length comes into view at a useful angle, and the two effects leave a slower falloff. Near the middle of any long rod (a few millimeters from a meter-long wire), the line result is an excellent approximation.

A long uniformly charged cylinder, solid or hollow, can be cut into thin lines parallel to its axis. At a point outside it, those lines together produce the same field as one line on the axis carrying the cylinder's charge per length $\lambda$, so $E = \frac{\lambda}{2\pi\varepsilon_0 r}$ with $r$ measured from the axis. Topic 8.6 gets that result, and the field inside the cylinder, from Gauss's law with no integral at all.

Arcs and rings: curved charge

When the charge lies on a circle, polar angle is the natural variable. A piece of arc of radius $R$ spanning angle $d\theta$ has length $R\,d\theta$, so $dq = \lambda R\,d\theta$.

At the center of an arc, every piece is the same distance $R$ away, which is what makes this case easy. Put the arc's center at the origin and let the arc run from $\theta = -\alpha$ to $\theta = +\alpha$, symmetric about the $+x$-axis. The piece at angle $\theta$ sits at $(R\cos\theta, R\sin\theta)$, and for positive charge its field at the center points from the piece toward the center, in the direction $(-\cos\theta, -\sin\theta)$. The $y$-components cancel in pairs, and

For a semicircle, $\alpha = \frac{\pi}{2}$ and $\lambda = \frac{Q}{\pi R}$, so the field at the center has magnitude

pointing away from the arc's midpoint for $Q > 0$. That is $\frac{2}{\pi} \approx 0.64$ of the $\frac{Q}{4\pi\varepsilon_0R^2}$ you'd get by pretending all the charge sat at one spot $R$ away, because the pieces near the ends of the semicircle push almost sideways and cancel each other. For a full circle, $\alpha = \pi$ and $\sin\pi = 0$: every piece has an opposite partner, and the field at the center of a uniform ring is zero.

On the axis of a ring, the field is not zero. Every piece of the ring is the same distance $\sqrt{x^2 + R^2}$ from a point a distance $x$ along the axis, and opposite pieces cancel each other's components perpendicular to the axis, but the components along the axis all point the same way and add. Example 3 carries out that integral and finds where the field is strongest.

These are the arrangements the exam expects you to integrate. Don't memorize the table: the exam rewards the setup (the $dq$, the $r$, the component, the limits), and each row takes a few lines once the setup is right. A nonuniform density changes only the $dq$: on a rod with $\lambda = \beta x$, use $dq = \beta x\,dx$ inside the same integral.

Worked examples

Worked exampleExample 3: the field along a ring's axis (free-response style)

A thin ring of radius $R$ carries charge $+Q$ spread uniformly. The ring lies in the $yz$-plane, centered on the origin, so the $x$-axis is its axis. Express answers in terms of $Q$, $R$, $x$ and physical constants.

(a) Derive an expression for the electric field at a point on the axis a distance $x$ from the center.

(b) At what distance from the center is the field strongest?

(c) Sketch $E_x$ as a function of $x$ from $x = -3R$ to $x = 3R$.

(d) A student estimates the field at $x = 3R$ by treating the ring as a point charge at the origin. Is the estimate too large or too small? Justify your answer, and find the percent difference.

Show solution

(a) Take a piece $dq$ of the ring. Its distance to $P$ is $r = \sqrt{x^2 + R^2}$, the same for every piece. The piece diametrically opposite has the same distance and a field whose component perpendicular to the axis is equal and opposite, so those components cancel around the ring and only $dE_x = dE\cos\theta$ survives, with $\cos\theta = \frac{x}{r}$:

$$E_x = \int\frac{1}{4\pi\varepsilon_0}\frac{dq}{r^2}\cdot\frac{x}{r} = \frac{x}{4\pi\varepsilon_0(x^2 + R^2)^{3/2}}\int dq = \frac{1}{4\pi\varepsilon_0}\frac{Qx}{(x^2 + R^2)^{3/2}},$$

directed along the axis away from the ring ($+x$ for $x > 0$, $-x$ for $x < 0$).

(b) Set the derivative to zero for $x > 0$:

$$\frac{dE_x}{dx} = \frac{Q}{4\pi\varepsilon_0}\cdot\frac{(x^2 + R^2) - 3x^2}{(x^2 + R^2)^{5/2}} = 0 \quad\Rightarrow\quad x = \frac{R}{\sqrt{2}}.$$

The derivative is positive for $x < \frac{R}{\sqrt{2}}$ and negative beyond, so this is a maximum, with $E_{\max} = \frac{2}{3\sqrt{3}}\cdot\frac{Q}{4\pi\varepsilon_0R^2} \approx 0.385\,\frac{Q}{4\pi\varepsilon_0R^2}$. By symmetry the field has the same magnitude at $x = -\frac{R}{\sqrt{2}}$.

(c) The graph is zero at the center, rises steeply to its peak at $x = 0.71R$, then decays toward zero, approaching the point-charge curve $\frac{Q}{4\pi\varepsilon_0x^2}$. For negative $x$ it is the same shape flipped below the axis, because the field points in $-x$ there.

(d) Too large. The exact field at $3R$ is $\frac{Q}{4\pi\varepsilon_0}\cdot\frac{3R}{(10R^2)^{3/2}} = 0.0949\,\frac{Q}{4\pi\varepsilon_0R^2}$, while the point-charge estimate is $\frac{Q}{4\pi\varepsilon_0(3R)^2} = 0.111\,\frac{Q}{4\pi\varepsilon_0R^2}$, about 17% larger. Two effects make the real field smaller: every piece of the ring is $\sqrt{10}R \approx 3.16R$ away rather than $3R$, and only the axial component, a fraction $\frac{3}{\sqrt{10}}$ of each piece's field, counts.

Where students lose points

On the exam

Derivations are graded on the setup as much as the answer: a correct $dq$ in terms of one variable ($\lambda\,dx$, $\lambda R\,d\theta$), the correct distance $r$, the surviving component with a stated reason the others cancel, and limits of integration that cover the object once. Give the direction with the magnitude, and expect to explain a result through a limit ("far away it matches a point charge") or to predict a factor of change from the result's dependences: a semicircle of fixed charge whose radius doubles has one quarter the field at its center, because $E \propto \frac{Q}{R^2}$. The exam's integrals here are limited to the rod, line, cylinder, ring and arc cases above; for other shapes, reason about direction from symmetry instead.