The Electric Field Due to a Point Charge — Question 2

Question 2

Two axial field readings determine an unknown source location

Use $k=8.99\times10^9\,\mathrm{N\,m^2/C^2}$; treat the specified charges as points in vacuum. One unknown point charge lies somewhere on the negative x axis. Its field at $x=0$ is $+360$ N/C and at $x=.200$ m is $+90.0$ N/C, both along x. No other field is present.

1. Infer the source sign and its distance left of the origin.

2. Calculate its charge and predict the field at $x=.400$ m.

Question 2 - Solution

Two axial field readings determine an unknown source location

(a) Method and calculation

$$Q>0,\quad\frac{360}{90}=\frac{(d+.200)^2}{d^2}=4\Rightarrow d=.200\,\mathrm m.$$$$Q=\frac{360(.200)^2}{k}\approx1.60178\,\mathrm{nC}.$$

(b) Interpretation and further calculation

$$E_x(.400)=\frac{kQ}{(.400+.200)^2}=40.0\,\mathrm{N/C}.$$$$x_Q=-.200\,\mathrm m.$$The decreasing positive field selects a positive source on the prescribed negative axis; that location restriction is part of the inference.

Physical check

Two magnitudes without directional or location information can permit additional source configurations.