Potential Due to a Continuous Charge Distribution — Question 4

Question 4

A finite rod approaches an infinite-line voltage difference without a finite absolute voltage

A uniform rod extends from x=-L to L with line density $\lambda>0$. Points lie a perpendicular distance b>0 from its midpoint. For finite L set zero potential at infinity.

1. Derive V(b).

2. Find the limit of V(b2)-V(b1) as L tends to infinity and explain why the individual voltages diverge.

Question 4 - Solution

A finite rod approaches an infinite-line voltage difference without a finite absolute voltage

(a) Method and calculation

$$V(b)=2k\lambda\int_0^L\frac{dx}{\sqrt{x^2+b^2}}=2k\lambda\operatorname{arsinh}(L/b).$$$$V(b)=2k\lambda\ln\left(\frac{L+\sqrt{L^2+b^2}}b\right).$$

(b) Interpretation and further calculation

$$L\gg b:\quad V(b)\simeq2k\lambda\ln(2L/b).$$$$\lim_{L\to\infty}[V(b_2)-V(b_1)]=2k\lambda\ln(b_1/b_2).$$

Physical check

The divergent term depends on L alone and cancels from the difference; the infinite-line model needs a finite-radius reference.