The Electric Field Due to a Point Charge — Question 1

Question 1

Displacement from the source fixes both the inverse-square factor and direction

Use $k=8.99\times10^9\,\mathrm{N\,m^2/C^2}$; treat the specified charges as points in vacuum. A source $Q=-5.00$ nC is at $(1,-2,0)$ m. Find its field at P, located at $(4,2,0)$ m.

1. Construct the displacement vector and calculate all field components.

2. Find the field at the point twice as far from the source on the same ray and the force there on a $+2.00$ nC probe.

Question 1 - Solution

Displacement from the source fixes both the inverse-square factor and direction

(a) Method and calculation

$$\mathbf r=\mathbf r_P-\mathbf r_Q=(3,4,0)\,\mathrm m,\quad r=5.00\,\mathrm m.$$$$\mathbf E=kQ\frac{\mathbf r}{r^3}=(-1.0788,-1.4384,0)\,\mathrm{N/C},\quad |\mathbf E|=1.798\,\mathrm{N/C}.$$

(b) Interpretation and further calculation

$$\mathbf E(2\mathbf r)=\mathbf E(\mathbf r)/4=(-.2697,-.3596,0)\,\mathrm{N/C}.$$$$\mathbf F=q\mathbf E=(-.5394,-.7192,0)\,\mathrm{nN}.$$The negative source reverses the outward displacement direction.

Physical check

Coordinates measured from the laboratory origin cannot replace displacement measured from the source.