Coulomb's Law — Question 4

Question 4

Perpendicular sources require signed vector components

Use $k=8.99\times10^9\,\mathrm{N\,m^2/C^2}$; treat the specified charges as points in vacuum. A $+1.00$ nC test charge is at the origin. Source A has $+2.00$ nC at $(.300,0)$ m; source B has $-3.00$ nC at $(0,.400)$ m.

1. Find the force vector on the test charge, including its signs and SI units.

2. Find the magnitude and direction measured counterclockwise from the positive x axis. Determine the external force needed to hold the test charge stationary.

Question 4 - Solution

Perpendicular sources require signed vector components

(a) Method and calculation

$$F_x=-k\frac{(1.00\times10^{-9})(2.00\times10^{-9})}{.300^2}=-1.99778\times10^{-7}\,\mathrm N.$$$$F_y=+k\frac{(1.00\times10^{-9})(3.00\times10^{-9})}{.400^2}=+1.68563\times10^{-7}\,\mathrm N.$$

(b) Interpretation and further calculation

$$|\mathbf F|=\sqrt{F_x^2+F_y^2}\approx2.61390\times10^{-7}\,\mathrm N.$$$$\theta=180^\circ-\tan^{-1}(|F_y/F_x|)\approx139.844^\circ,\quad\mathbf F_{\mathrm{hold}}=-\mathbf F.$$The attractive vertical contribution points toward B; the repulsive horizontal contribution points away from A.

Physical check

The magnitude is smaller than $|F_x|+|F_y|$ because the two contributions are perpendicular.