Coulomb's Law — Question 2

Question 2

A force-free point between like charges is stable only along the joining line

Use $k=8.99\times10^9\,\mathrm{N\,m^2/C^2}$; treat the specified charges as points in vacuum. Fixed charges $+4.0$ nC and $+9.0$ nC lie at $x=0$ and $x=1.00$ m. A positive test charge $q$ is placed between them. Its influence on the fixed sources is neglected.

1. Find the position where the test charge force vanishes.

2. Differentiate the longitudinal force to determine stability for small displacements along the line. Explain the force direction for a small upward displacement from that point.

Question 2 - Solution

A force-free point between like charges is stable only along the joining line

(a) Method and calculation

$$F_x=kq\left(\frac{4.0\times10^{-9}}{x^2}-\frac{9.0\times10^{-9}}{(1-x)^2}\right)=0.$$$$\frac2x=\frac3{1-x}\Rightarrow x=.400\,\mathrm m.$$

(b) Interpretation and further calculation

$$\frac{dF_x}{dx}=-2kq\left(\frac{4.0\times10^{-9}}{x^3}+\frac{9.0\times10^{-9}}{(1-x)^3}\right)<0.$$$$\delta x>0\Rightarrow\delta F_x<0,\qquad\delta y>0\Rightarrow F_y>0.$$The longitudinal force is restoring, but both positive sources push an upward-displaced positive charge farther upward. This is not a stable unconstrained three-dimensional trap.

Physical check

For a negative test charge the force-free location is unchanged, but the local restoring and destabilizing directions reverse.