The Electric Field Due to a Point Charge — Question 9

Question 9

A distant pair resembles one charge with a calculable finite-size correction

Two equal positive point charges Q lie at $x=\pm a$. Observe their field at $x>a$. A point approximation replaces them by charge $2Q$ at the origin.

1. Derive the exact field and its ratio to the point approximation.

2. Evaluate the fractional excess over the point approximation at $x=10a$ and derive the leading far-field correction.

Question 9 - Solution

A distant pair resembles one charge with a calculable finite-size correction

(a) Method and calculation

$$E_{\mathrm{exact}}=kQ[(x-a)^{-2}+(x+a)^{-2}]=\frac{2kQ(x^2+a^2)}{(x^2-a^2)^2}.$$$$E_{\mathrm{point}}=2kQ/x^2,\quad\frac{E_{\mathrm{exact}}}{E_{\mathrm{point}}}=\frac{1+u^2}{(1-u^2)^2},\quad u=a/x.$$

(b) Interpretation and further calculation

$$u=.1:\quad E_{\mathrm{exact}}/E_{\mathrm{point}}-1\approx.0305071.$$$$\frac{1+u^2}{(1-u^2)^2}=1+3u^2+O(u^4).$$The leading error is quadratic in source size divided by distance because the symmetric first-order terms cancel.

Physical check

The fractional error here is explicitly normalized by the approximate field.