Question 23.1: The Electric Field Due to a Charged Rod A rod of length l ha...
The Electric Field Due to a Charged Rod
A rod of length \ell has a uniform positive charge per unit length λ and a total charge Q. Calculate the electric field at a point P that is located along the long axis of the rod and a distance a from one end (Fig. 23.2).
Conceptualize The field d\overrightarrow{E} at P due to each segment of charge on the rod is in the negative x direction because every segment carries a positive charge. Figure 23.2 shows the appropriate geometry. In our result, we expect the electric field to become smaller as the distance a becomes larger because point P is farther from the charge distribution.
Categorize Because the rod is continuous, we are evaluating the field due to a continuous charge distribution rather than a group of individual charges. Because every segment of the rod produces an electric field in the negative x direction, the vector sum of their contributions is easy to determine.
Analyze Let’s assume the rod is lying along the x axis, dx is the length of one small segment, and dq is the charge on that segment. Because the rod has a charge per unit length λ, the charge dq on the small segment is dq = λ dx.
Find the magnitude of the electric field at P due to one segment of the rod having a charge dq:
d E=k_e \frac{d q}{x^2}=k_e \frac{\lambda d x}{x^2}Find the total field at P using¹ Equation 23.1:
\overrightarrow{E}=k_e \lim _{\Delta q_i \rightarrow 0} \sum_i \frac{\Delta q_i}{r_i^2} \hat{r}_i=k_e \int \frac{d q}{r^2} \hat{r} (23.1)
E=\int_a^{\ell+a} k_e \lambda \frac{d x}{x^2}Noting that k_e and \lambda=Q / \ell are constants and can be removed from the integral, evaluate the integral:
E=k_e \lambda \int_a^{\ell+a} \frac{d x}{x^2}=k_e \lambda\left[-\frac{1}{x}\right]_a^{\ell+a}(1) E=k_e \frac{Q}{\ell}\left(\frac{1}{a}-\frac{1}{\ell+a}\right)=\frac{k_e Q}{a(\ell+a)}
Finalize We see that our prediction is correct; if a becomes larger, the denominator of the fraction grows larger, and E becomes smaller. On the other hand, if a → 0, which corresponds to sliding the bar to the left until its left end is at the origin, then E → ∞. That represents the condition in which the observation point P is at zero distance from the charge at the end of the rod, so the field becomes infinite. We explore large values of a below.
¹To carry out integrations such as this one, first express the charge element dq in terms of the other variables in the integral. (In this example, there is one variable, x, so we made the change dq = λ dx.) The integral must be over scalar quantities; therefore, express the electric field in terms of components, if necessary. (In this example, the field has only an x component, so this detail is of no concern.) Then, reduce your expression to an integral over a single variable (or to multiple integrals, each over a single variable). In examples that have spherical or cylindrical symmetry, the single variable is a radial coordinate.