Let's look at an example. $$\let\lm\lambda$$

Caption

Let's say that the line lies along the \(x\)-axis with the origin at the center. The target (where we want to find the electric field) is on the positive \(y\)-axis. To find the electric field there, we break the line up into a bunch of little charges \(dq\) and find the field due to each one.

For instance, let's consider the little piece of the line highlighted in red, which is a distance \(x\) from the origin. The field it creates at the star is $$d\vec E=k{dq\over d^3}\vec d$$ To put this into integrable form, we need to figure out what \(dq\) and \(\vec d\) are.

- Just as in the previous section, we need to rewrite \(dq\) in terms of a variable we can actually integrate over. The charge of this segment is equal to its line charge density \(\lm(x)\) times its length \(dL\), and we can write its length as \(dx\) because it lies along the \(x\) axis. Therefore $$dq=\lm(x)\,dx$$
- The vector \(\vec d\) points from source to target, as shown. We want to write this in terms of its components, and we do that by asking "How can we get from the source to the target by only moving horizontally or vertically?" In this case, we need to move upward a distance \(H\), and then a distance \(x\) to the left (or vice versa). Thus $$\vec d=H\hat y-x\hat x$$ We'll also need the length of this vector, which is $$d=\sqrt{H^2+x^2}$$

$$\int \frc{(H^2+x^2)^{3/2}}\,dx \let\th\theta$$

Let \(x=H\tan\th\) and so \(dx=H\sec^2\th\,d\th\).
$$\begin{eqnarray}
&=&\int\frac{H\sec^2\th\,d\th}{H^3(1+\tan^2\th)^{3/2}}\\
&=&\frc{H^2}\int\frac{\sec^2\th\,d\th}{(\sec^2\th)^{3/2}}\\
&=&\frc{H^2}\int\frac{d\th}{\sec\th}\\
&=&\frc{H^2}\int\cos\th\,d\th\\
&=&\frc{H^2}\sin\th\\
&=&\frc{H^2}{\tan\th\over\sec\th}\\
&=&\frc{H^2}{\tan\th\over \sqrt{1+\tan^2\th}}\\
&=&\frc{H^2}{x/H\over \sqrt{1+x^2/H^2}}\\
\end{eqnarray}$$

$$={x\over H^2\sqrt{H^2+x^2}}$$
$$\int\frac{x\,dx}{(H^2+x^2)^{3/2}}$$

Let \(y=x^2+H^2\) so \(dy=2x\,dx\).
$$\begin{eqnarray}
&=&\int\frac{\frc2\,dy}{y^{3/2}}\\
&=&\frc2\int y^{-3/2}\,dy\\
&=&\frc2(-2) y^{-1/2}\\
\end{eqnarray}$$

$$=-\frc{\sqrt{H^2+x^2}}$$
The electric field of a line charge of length \(L\).

- To be "really far" from the line means that \(H\gg L\), so that \(H^2+(L/2)^2\approx H^2\). Thus the electric field far from the line is $$\vec E\approx {k\lm L\over H\sqrt{H^2}}\hat y={k Q\over H^2}\hat y,\,(H\gg L)$$ where \(Q=\lm L\) is the total charge of the line. Because \(H\) is the distance of the target from the line and \(\hat y\) is the unit vector that points from the line to the target, we might write this as \(\vec E={kQ\over r^2}\hat r\), which is the electric field of a point. Why? Well, we can also interpret \(H\gg L\) as \(L\ll H\)--that is, the line is really small. A very small line is basically a point, so its field will be the field of a point charge. (The field of any finite charge distribution will do the same thing if you get far enough away from it.)
- To be "really close" to the line means that \(H\ll L\) and \(H^2+(L/2)^2 \approx (L/2)^2\). $$E\approx {k\lm L\over H\sqrt{(L/2)^2}}\hat y={2k\lm\over H}\hat y ,\, (H\ll L)$$ This is also the field of a very long line. Notice that the field doesn't depend on the length \(L\): once the line is "long enough", adding more charge to either end of the line doesn't affect the electric field much. In fact, this is also the field of an infinite line (take the limit \(L\to\infty\)). Notice that the field dies away as \(1/r\) rather than \(1/r^2\): this is because the line is not finite, so that no matter how far away you get from it, it never looks like a point. Any charge distribution which is infinite in one dimension will have this same property that \(E\sim \frc{r}\).