Question

-6 points osuriPhys1 22 3.P Two charges, +6 με and +15 μc, are fixed 1 m apart, with the second one to the right. Find the magnitude and direction of the net force in N on a-5 nC charge when placed at the following locations. 7. 4 43 My Notes (a) helfway between the two magnitude direction (b) half a meter to the left of the +6 pC charge IN magnitude direction () helf a meter above the +15 uC charge in a direction perpendicular to the line joining the two ixed charges (Assume this line is the x-axis with the +x-direction toward the right. Indicate the direction of the force in degrees counterclockwise from the x axis.) magnitude direction counterclockwise from the +x-axis Additional Materials eBook

Answer 1

F^rightarrow = F^rightarrow_1 + F^rightarrow_2 F^rightarrow = [- 9 times 10^9 times 6 times 10^-6 times 5 times 10^-9/(0.5)^2 + 9 times 10^9 times 15 times 10^-6 times 5 times 10^-9/(0.5)^2] N rightarrow +ve directions F^rightarrow = 1.62 times 10^-3 N t therefore magnitude [1.62 times 10^-3] N Direction [towards right] F^rightarrow F^rightarrow_1 + F^rightarrow_2 F^riaghtarrow = 9 times 10^9 5 times 10^-9 [6 times^-6/(0.5)^2 + 15 times 10^-6/(0.5)^2] N therefore F^rightarrow = 1380 times 10^-6 N = 1.3 times 10^-3 N uparrow therefore magnitude [1.380 times 10^-3] N Direction [towards right] F^rightarrow = F^rightarrow_1 + F^rightarrow_2 F^rightarrow = - Vertical-bar F_1 Vertical-bar 1/ squareroot 1.25 uparrow + [ Vertical-bar F_1 Vertical-bar 0.5/squareroot 1.25 - Vertical-bar F_2 Vertical-bar ] uparrow Vertical-bar F_1 Vertical-bar = 9 times 10^+ 9 times 5 times 10^-9 times 6 times 10^-6/(squareroot 1.25)^2 N = 2.16 times 10^-4 N Vertical-bar F_2 Vertical-bar = 9 times 10^-9 times 5 times 10^-9 times 15 times 10^-6/(0.5)^2 N = 27.0 times 10^-4 N F^rightarrow = -1.93 times 10^-4 N uparrow - 27.966 times 10^-4 N uparrow therefore magnitude = [2.80 times 10^-3] N