Question

PART A). Charge Q1=6.0 nC is at (0.30 m, 0), charge Q2=−1.0 nC is at (0, 0.10 m), and charge Q3=5.0 nC is at (0, 0). What is the magnitude of the net electrostatic force on the 5.0-nC charge due to the other charges? (k=1/4πϵ0=8.99×109 N · m2/C2). PART B). What is the direction of the net electrostatic force on the 5.0-nC charge due to the other charges?

Answer 1

Tools Screen 1 of 1 E View Options Close The magnitude of the net electrostatic force is just the sum of the forces that act on Q3. However, in this case, the forces is perpendicular to each other, one is in y-axis and the other in x-axis, therefore, we have to use Pythagorean Theorem to add them Force Q1 on Q3 F13-9"10ng * 6*10л-9 * 5*10л-9/(/.3-0)"2 Force Q2 on Q3 F23 = 9*10ng *-1*10^-9 * 5* 10л-9/(1-0)12 F23-4.5*1 0n_6N Fnet Sqrt((3-10л-6)^2+(-4.5*10л-6)^2) Fnet -5.4*10A-6N ENG US 00:39 23-07-2016 n !''' .וזן' )