The Electric Field Due to a Line of Charge — Question 3

Question 3

Near an endpoint a uniformly charged rod cannot be replaced by its center charge

A uniform rod occupies $0\le x\le L$ with $L=.400$ m and $\lambda=3.00$ nC/m. Observe at $x=-s$, where $s=.200$ m. Use $k=8.99\times10^9$ N m$^2$/C$^2$.

1. Integrate the axial field and evaluate it.

2. Compare with placing the total rod charge at its midpoint; report the fractional underestimate relative to the exact magnitude.

Question 3 - Solution

Near an endpoint a uniformly charged rod cannot be replaced by its center charge

(a) Method and calculation

$$E_x=-k\lambda\int_0^L\frac{dx}{(s+x)^2}=-k\lambda\left(\frac1s-\frac1{s+L}\right).$$$$E_x=-89.9\,\mathrm{N/C},\quad Q=\lambda L=1.20\,\mathrm{nC}.$$

(b) Interpretation and further calculation

$$E_{\mathrm{center}}=-\frac{kQ}{(s+L/2)^2}=-67.425\,\mathrm{N/C}.$$$$\frac{|E_{\mathrm{exact}}|-|E_{\mathrm{center}}|}{|E_{\mathrm{exact}}|}=.250.$$The closer part of the rod contributes disproportionately because the kernel varies as inverse distance squared.

Physical check

The field points left, away from every positive source element.