MODEL The rod is thin, so we’ll assume the charge lies along a line and forms what we call a line of charge. The rod’s linear charge density is $\text{λ}=Q\text{/}L$.

VISUALIZE FIGURE 23.12 illustrates the steps of the problem-solving strategy. Distance r is simply a distance; we need a coordinate system to carry out the calculation. We’ve chosen a coordinate system in which the rod lies along the y-axis and point P, in the bisecting plane, is at distance x on the x-axis. We’ll calculate the field at this point and then, at the end, replace x with the more generic r. We’ve then divided the rod into N small segments of charge $\text{Δ}Q$, each of which is small enough to model as a point charge. For every $\text{Δ}Q$ in the bottom half of the wire with a field that points to the right and up, there’s a matching $\text{Δ}Q$ in the top half whose field points to the right and down. The y-components of these two fields cancel, hence the net electric field on the x-axis points straight away from the rod. The only component we need to calculate is $E_x$.

FIGURE 23.12 Calculating the electric field of a line of charge.

SOLVE Each of the little segments of charge can be modeled as a point charge. We know the electric field of a point charge, so we can write the x-component of $\vec{E}{}_i$, the electric field of segment i, as

where $r_i$ is the distance from charge i to point P. You can see from the figure that $r_i={\left(y_i^2+x^2\right)}^{1\text{/}2}$ and $\text{cos }{\theta }_i=x\text{/}{r}_i=x\text{/}{\left(y_i^2+x^2\right)}^{1\text{/}2}$. With these, ${\left(E_i\right)}_x$ is

If we now sum this expression over all the charge segments, the net x-component of the electric field is

This is the same superposition we did for the $N=3$ case in Example 23.1. The only difference is that we have now written the result as an explicit summation so that N can have any value. We want to let $N\to \infty$ and to replace the sum with an integral, but we can’t integrate over Q; it’s not a geometric quantity. This is where the linear charge density enters. The quantity of charge in each segment is related to its length $\text{Δ}y$ by $\text{Δ}Q=\text{λ}\text{Δ}y=\left(Q\text{/}L\right)\text{Δ}y$. In terms of the linear charge density, the electric field is

Now we’re ready to let the sum become an integral. If we let $N\to \infty$, then each segment becomes an infinitesimal length $\text{Δ}y\to dy$ while the discrete position variable $y_i$ becomes the continuous integration variable y. The sum from $i=1$ at the bottom end of the line of charge to $i=N$ at the top end will be replaced with an integral from $y=-L\text{/}2$ to $y=+L\text{/}2$. Thus in the limit $N\to \infty$,

This is a standard integral that you have learned to do in calculus and that can be found in Appendix A. Note that x is a constant as far as this integral is concerned. Integrating gives

The field depends on only the distance r from the wire, not on a specific coordinate system, so we can now replace x with r to write that the electric field strength $E_{\text{rod}}$ at distance r from the center of a charged rod is

The field strength must be positive, so we added absolute value signs to Q to allow for the possibility that the charge could be negative. The only restriction is to remember that this is the electric field at a point in the plane that bisects the rod.

REVIEW Suppose we are at a point very far from the rod. If $r\gg L$, the rod appears to be a point charge Q in the distance. Thus in the limiting case $r\gg L$, we expect the rod’s electric field to be that of a point charge. If $r\gg L$, the square root becomes $(r^2+{\left(L\text{/}2\right)}^2{)}^{1\text{/}2}\approx {\left(r^2\right)}^{1\text{/}2}=r$ and the electric field strength at distance r becomes $E_{\text{rod}}\approx Q\text{/}4\pi {ϵ}_0{r}^2$, the field of a point charge. The fact that our expression of $E_{\text{rod}}$has the correct limiting behavior gives us confidence that we haven’t made any mistakes in its derivation.