You know how to add the fields of several point charges: find each one's field at the point, then add them as vectors. A charged rod is the same problem with infinitely many charges. Cut it into pieces $dq = \lambda\,dx$ (charge densities), treat each piece as a point charge, and add the pieces with an integral.
Key idea
Each piece makes $dE = \dfrac{k\,dq}{r^2}$, pointing away from the piece (for positive charge). Split it into components and integrate each one separately:
$$E_x = \int dE_x, \qquad E_y = \int dE_y.$$
Use symmetry first: components that cancel in mirror-image pairs integrate to zero.
Why it works
An integral adds numbers, not arrows. Two arrows of size 3 at right angles add to 4.2, not 6. So we turn one vector sum into two ordinary sums, one per component, where plain addition is correct. The size $dE$ may be added directly only when every piece's field points the same way, as for a point on the rod's own line.
A recipe for any shape
- Draw axes and one piece $dq$ at a general position, say $x$.
- Write its distance $r$ to the point, and the angle or ratio that gives each component (for example $\cos\theta = d/r$).
- Write $dE_x$ and $dE_y$ in terms of $x$ and $dx$.
- Use symmetry to drop components that cancel.
- Integrate over the whole object, then check a limit you know (far away, it must look like a point charge).
Worked examples
Worked example: On the rod's own line
A rod $0.20\ \text{m}$ long carries $8.0\ \text{nC}$ spread evenly. Find the field at a point on the rod's line, $0.10\ \text{m}$ beyond one end.
- Put P at the origin and the rod from $x = a = 0.10\ \text{m}$ to $x = a + L = 0.30\ \text{m}$. Every piece pushes along the same line, so here the sizes may be added.
- $$E = \int_a^{a+L}\frac{k\lambda\,dx}{x^2} = k\lambda\left(\frac{1}{a} - \frac{1}{a+L}\right) = \frac{kQ}{a(a+L)}$$ using $\lambda L = Q$.
- $$E = \frac{(8.99 \times 10^{9})(8.0 \times 10^{-9})}{(0.10)(0.30)} = 2400\ \text{N/C}$$
- Check: far away ($a \gg L$) this is $kQ/a^2$, a point charge.
Answer: $E \approx 2400\ \text{N/C}$, along the rod's line, away from the rod
Worked example: Above the middle
A rod $0.40\ \text{m}$ long carries $10\ \text{nC}$. Find the field $d = 0.10\ \text{m}$ above its midpoint.
- Rod from $x = -L/2$ to $L/2$, P at $(0, d)$. A piece at $x$ is $r = \sqrt{x^2 + d^2}$ away.
- Mirror pieces at $\pm x$ cancel each other's $x$-parts. The $y$-part of each is $dE\cos\theta = \dfrac{k\lambda\,dx}{r^2}\cdot\dfrac{d}{r}$.
- $$E = k\lambda d\int_{-L/2}^{L/2}\frac{dx}{(x^2 + d^2)^{3/2}}$$ With the substitution $x = d\tan\theta$, $dx = d\sec^2\theta\,d\theta$, the integrand becomes $\cos\theta\,d\theta/d^2$, so the antiderivative is $\dfrac{x}{d^2\sqrt{x^2 + d^2}}$.
- $$E = \frac{k\lambda L}{d\sqrt{d^2 + L^2/4}} = \frac{kQ}{d\sqrt{d^2 + L^2/4}}$$
- $$E = \frac{(8.99 \times 10^{9})(10 \times 10^{-9})}{(0.10)\sqrt{0.010 + 0.040}} = 4020\ \text{N/C}$$ pointing straight away from the rod.
Answer: $E \approx 4020\ \text{N/C}$
Worked example: Two checks on the answer
Check the result $E = kQ/\left(d\sqrt{d^2 + L^2/4}\right)$ far away and very close.
- Far away, $d \gg L$: the square root is about $d$, so $E \approx kQ/d^2$. From far off, the rod looks like a point.
- Very close, $d \ll L$: the root is about $L/2$, so $E \approx \dfrac{2kQ}{dL} = \dfrac{2k\lambda}{d}$. Up close, the rod looks endless. Gauss's law will give the same $2k\lambda/d$ in one line.
Answer: Both limits come out right, a good sign the integral is correct.
Common mistake
Adding the sizes $dE$. Integrating $k\lambda\,dx/r^2$ without the factor $d/r$ treats every piece's field as if it pointed straight up. For the second example that gives $\dfrac{2k\lambda}{d}\arctan\dfrac{L}{2d} = 4980\ \text{N/C}$, about 24% too big. The sideways parts point left and right and cancel; only the parts along the bisector survive. Quick check: unless every piece pushes along the same line, the true field is smaller than the sum of the sizes.
At a point that is not on a line of symmetry (above one end of the rod, say), nothing cancels. Then you must integrate both $dE_x$ and $dE_y$, and the field tilts away from the rod.