q1 = -4.00nC @ (0.6m,0.8m) q2 = +6.00nC @ (0.6m, 0.0m) Enet = 132 N/C Direction
ID: 2133992 • Letter: Q
Question
q1 = -4.00nC @ (0.6m,0.8m)
q2 = +6.00nC @ (0.6m, 0.0m)
Enet = 132 N/C
Direction (Theta) = ?
A point charge = -4.00 is at the point x = 0.60m , y= 0.80m, and a second point charge q2 = +6.00 nc is at the point x= 0.60m, y= 0.00m. What is the magnitude of the net electric field on the origin (0,0) and the direction of the net electric field? Already hit give up on the magnitude after 1.5 hours of wrong answers and the answer came out to be 132 N/C, not sure how it got that answer but I was close with a few of my answers. Just want to know what the direction is supposed to be? Thanks!Explanation / Answer
E = kq/r^2
From the first charge
E = (9 X 10^9)(4 X 10^-9)/1 = 36 N/C
The will have x and y components
tan(angle) = .8/.6
angle = 53.1 degrees
Ex = (36)(cos 53.1) = 21.6 N/C upward
Ey = (36)(sin 53.1) = 28.8 N/C to the right
For the second charge
E = (9 X 10^9)(6 X 10^-9)/(.6)^2
E = 150 N/C to the left
The net right/left = 150 - 28.8 = 121.2 to the left
The net up/down is 21.6 up
The net E is found from the pythagorean theorem
Net = sqrt[(21.6)^2 + (121.2)^2]
Net = 123 N/C
The angle is found from the tangent function
tan(angle) = 21.6/121.1
angle = 10.1 degrees North of West
That is 10.1 degrees above the negative x axis or 169.9 degrees from the positive x axis
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