Potential Due to a Point Charge — Question 3

Question 3

Two voltages along one ray determine a hidden source position

A positive point charge lies on the x axis at an unknown $x_s<0$. With V(infinity)=0, measurements are V(0)=90.0 V and V(d)=60.0 V, where d=.200 m.

1. Find $x_s$ without first determining Q.

2. Find kQ and predict V(2d).

Question 3 - Solution

Two voltages along one ray determine a hidden source position

(a) Method and calculation

$$a=-x_s>0,\quad\frac{V(0)}{V(d)}=\frac{a+d}{a}=\frac32.$$$$a=2d=.400\,\mathrm m,\quad x_s=-.400\,\mathrm m.$$

(b) Interpretation and further calculation

$$kQ=V(0)a=36.0\,\mathrm{V\,m}.$$$$V(2d)=\frac{36.0}{.400+.400}=45.0\,\mathrm V.$$

Physical check

The specified side of the source removes the absolute-distance ambiguity of a two-point measurement.