The Electric Field Due to a Line of Charge — Question 5

Question 5

An antisymmetric density makes a neutral rod act as a dipole at large distance

A thin rod on $[-L,L]$ has signed density $\lambda(x)=\alpha x$, where $\alpha>0$ has units C/m$^2$. Observe at $(0,y)$, $y>0$. You may use $\int dx/\sqrt{x^2+y^2}=\ln(x+\sqrt{x^2+y^2})$.

1. Derive the field components by parity and integration.

2. Find the total charge and dipole moment, then state the leading far-field vector using the dipole result.

Question 5 - Solution

An antisymmetric density makes a neutral rod act as a dipole at large distance

(a) Method and calculation

$$E_y=k\alpha y\int_{-L}^L\frac{x\,dx}{(x^2+y^2)^{3/2}}=0,\quad E_x=-2k\alpha\int_0^L\frac{x^2\,dx}{(x^2+y^2)^{3/2}}.$$$$E_x=-2k\alpha\left[\ln\frac{L+\sqrt{L^2+y^2}}y-\frac{L}{\sqrt{L^2+y^2}}\right].$$

(b) Interpretation and further calculation

$$Q=\alpha\int_{-L}^Lx\,dx=0,\quad p_x=\alpha\int_{-L}^Lx^2\,dx=\frac{2\alpha L^3}{3}.$$$$y\gg L:\quad\mathbf E\simeq-\frac{2k\alpha L^3}{3y^3}\hat{\mathbf x}.$$Opposite density signs change which components cancel compared with a uniformly positive rod.

Physical check

The density is deliberately signed; it is not a positive-charge density that becomes unphysical on half the rod.