Net Force on Charged particle
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Net Force on Charged particle

[From: ] [author: ] [Date: 12-07-11] [Hit: ]
y) coordinate system. An additional +5.33 C charge is located at (4.00 m,0.00) and –2.......
A 1.91 C charge is located at the origin of an (x,y) coordinate system. An additional +5.33 C charge is located at (4.00 m,0.00) and –2.91 C charge is located at (0.00,7.00 m).

a) What is the magnitude of the net force on the 1.91 C charge at the origin?


b) What is the direction of the net force on the 1.91 C charge at the origin (with respect to the positive x-axis)?

-
F = Kc q1 q2 / r^2
F1 = 8.99E9 (1.91 * 5.33) / 16 = 5.72E9 N at 180 deg (to left on x axis - repulsion)
F2 = 8.99E9 (1.91 * -2.91) / 49 = 1.02E9 N at 90 deg (up on y axis - attraction)
draw the two vectors
construct parallelogram
find diagonal (resultant)
I get
5.81 N at 169.9 deg or 5.81 N at 10.1 deg up from neg x axis
1
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