Coulomb's Law — Question 3

Question 3

Opposite source signs move the cancellation point outside the interval

Use $k=8.99\times10^9\,\mathrm{N\,m^2/C^2}$; treat the specified charges as points in vacuum. A charge $+4.0$ nC is fixed at $x=0$ and $-1.0$ nC at $x=.600$ m. Locate a finite position on the axis where the force on a nonzero test charge vanishes.

1. Eliminate the interval between the charges and the region $x<0$ using force directions and relative magnitudes.

2. Solve in $x>.600$ m and verify the result using both distances. Explain why equating magnitudes alone can produce a spurious candidate.

Question 3 - Solution

Opposite source signs move the cancellation point outside the interval

(a) Method and calculation

$$0<x<.600:\quad\text{both forces on a positive test charge point right}.$$$$x<0:\quad\frac{F_{+4}}{F_{-1}}=4\frac{(.600-x)^2}{x^2}>4,\quad F_x\ne0.$$

(b) Interpretation and further calculation

$$x>.600:\quad\frac4{x^2}=\frac1{(x-.600)^2}\Rightarrow\frac2x=\frac1{x-.600}.$$$$x=1.200\,\mathrm m,\quad\frac4{1.200^2}=\frac1{.600^2}.$$At this location the forces oppose. A squared magnitude equation cannot establish that directional requirement by itself.

Physical check

A test charge of either sign has zero force at the same point because its sign multiplies both contributions.