Potential Due to a Continuous Charge Distribution — Question 3

Question 3

An axial rod potential contains a logarithm of two endpoint distances

A uniform rod occupies $0\le x\le L$ and has positive line density $\lambda$. Observe at x=-a with a>0. Use infinity reference.

1. Integrate the potential and express the logarithm using a dimensionless ratio.

2. Find its leading approximation for a much larger than L and the first relative correction to the point-charge result.

Question 3 - Solution

An axial rod potential contains a logarithm of two endpoint distances

(a) Method and calculation

$$V(-a)=k\lambda\int_0^L\frac{dx}{a+x}=k\lambda[\ln(a+x)]_0^L.$$$$V(-a)=k\lambda\ln(1+L/a).$$

(b) Interpretation and further calculation

$$V=\frac{k\lambda L}{a}\left[1-\frac{L}{2a}+O(L^2/a^2)\right].$$$$Q=\lambda L,\quad\frac{V-kQ/a}{kQ/a}\simeq-\frac L{2a}.$$

Physical check

The expansion uses the near endpoint as origin; choosing the rod center would remove its first-order displacement correction.