Electric Fields via Integration

To find the electric field created by a charge distribution, we use the same basic technique: break the distribution up into a bunch of little pieces, find the electric field at the target due to each little piece, and adding all the fields together. We can think of each little piece as a little point charge \(dq\), and the field it creates as $$d\vec E=k{dq\over d^3}\vec d$$ where \(\vec d\) is the vector from the source to the target. (Notice that this vector isn't necessarily tiny, which is why we don't write it as an infinitesimal. The electric field created by a tiny point charge is tiny, however, so I write \(d\vec E\). The total electric field is then $$\vec E=\int d\vec E=\int k{dq\over d^3}\vec d$$

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

Electric Field of a Line Charge


Caption
Suppose we have a line charge of length \(L\) and line charge density \(\lambda(x)\). Find the electric field a distance \(H\) from the center of the line.

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.

Now it may seem that we only found the electric field due to one particular piece of the line, but by changing the value of x we can find the field due to any of the sources. Putting all the pieces together, we get $$\vec E=\int d\vec E=\int_{-L/2}^{L/2} k{\lambda(x)\,dx\over (H^2+x^2)^{3/2}}(H\hat y-x\hat x)$$ This is the integral of a vector, and you may be wondering how we're supposed to handle that. The answer is simple: the unit vectors \(\hat x\), \(\hat y\), and \(\hat z\) are all constants: they always point in their respective directions no matter what \(x\) is. Therefore, we can pull them out of the integral: $$\vec E=kH\hat y\int_{-L/2}^{L/2} {\lambda(x)\,dx\over (H^2+x^2)^{3/2}} -k\hat x\int_{-L/2}^{L/2} {x\lambda(x)\,dx\over (H^2+x^2)^{3/2}}$$
$$\int \frc{(H^2+x^2)^{3/2}}\,dx \let\th\theta$$
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 \(dy=2x\,dx\). $$\begin{eqnarray} &=&\int\frac{\frc2\,dy}{y^{3/2}}\\ &=&\frc2\int y^{-3/2}\,dy\\ &=&\frc2(-2) y^{-1/2}\\ \end{eqnarray}$$
$$=-\frc{\sqrt{H^2+x^2}}$$

Constant charge density

Without knowing what \(\lambda(x)\) is, this is the best we can do. The simplest case is if it is a constant \(\lm\) and can come out of the integral. We can then use the two integral identities shown on the left and right. Then the electric field is $$\begin{eqnarray} \vec E&=& kH\lm\hat y \int_{-L/2}^{L/2}{dx\over (H^2+x^2)^{3/2}}-k\lm\hat x\int_{-L/2}^{L/2}{x\,dx\over (H^2+x^2)^{3/2}}\\ &=&kH\lm\hat y\left[{x\over H^2\sqrt{H^2+x^2}}\right]_{-L/2}^{L/2} -k\lm \hat x\left[-\frc{\sqrt{x^2+H^2}}\right]_{-L/2}^{L/2}\\ &=& {k\lm \hat y\over H}\left[ {L/2\over\sqrt{H^2+(L/2)^2}}-{-L/2\over\sqrt{H^2+(-L/2)^2}}\right]\\ &&+k\lm\hat x\left[\frc{\sqrt{H^2+(L/2)^2}}-\frc{\sqrt{H^2+(-L/2)^2}}\right]\\ &=&{k\lm L\over H\sqrt{H^2+(L/2)^2}}\hat y+0\\ \end{eqnarray}$$ The second integral is zero because the integrand is an odd function of \(x\), so the electric field at the star points in the \(\hat y\) direction, as one might expect from symmetry.

The electric field of a line charge of length \(L\).

Limiting cases

We can graph the electric field as a function of \(H/L\) as shown, but this doesn't tell us much other than that the electric field gets weaker as one moves away from the line. We can gain more insight if we consider what happens when one is really close to the line, and when one is really far from the line. (Read Section 0.2 on approximations if you need a refresher.)