Coulomb's Law — Question 5

Question 5

Opposite base charges cancel vertical forces at an equilateral apex

Use $k=8.99\times10^9\,\mathrm{N\,m^2/C^2}$; treat the specified charges as points in vacuum. An equilateral triangle of side $a=.200$ m has $+q$ at the left base corner, $-q$ at the right base corner, and $+q$ at the top, where $q=2.00$ nC. The base is horizontal and all charges are fixed.

1. Resolve the two forces on the top charge and find its resultant.

2. Replace the right base charge by $+q$ and find the new magnitude and direction. Compare the two results without treating signed charge as a negative force magnitude.

Question 5 - Solution

Opposite base charges cancel vertical forces at an equilateral apex

(a) Method and calculation

$$F_0=kq^2/a^2=8.99\times10^{-7}\,\mathrm N.$$$$\mathbf F_L=F_0(\tfrac12\hat{\mathbf i}+\tfrac{\sqrt3}2\hat{\mathbf j}),\quad\mathbf F_R=F_0(\tfrac12\hat{\mathbf i}-\tfrac{\sqrt3}2\hat{\mathbf j}).$$$$\mathbf F_{\mathrm{top}}=F_0\hat{\mathbf i}.$$

(b) Interpretation and further calculation

$$\mathbf F_R\prime=F_0(-\tfrac12\hat{\mathbf i}+\tfrac{\sqrt3}2\hat{\mathbf j}).$$$$\mathbf F_{\mathrm{top}}\prime=\sqrt3F_0\hat{\mathbf j}\approx1.55711\times10^{-6}\hat{\mathbf j}\,\mathrm N.$$Changing one source sign exchanges vertical cancellation for horizontal cancellation.

Physical check

The fixed supports supply the forces needed to prevent motion; a nonzero resultant is consistent with charges held in place.