Three point charges are arranged as shown in the figure. (Take q1 = 6.54 nC, q2
ID: 2005530 • Letter: T
Question
Three point charges are arranged as shown in the figure. (Take q1 = 6.54 nC, q2 = 5.35 nC, and q3 = -3.24 nC.)
(a) Find the magnitude of the electric force on the particle at the origin. Answer in N.
(b) Find the direction of the electric force on the particle at the origin. (counter-clockwise from the +x-axis).
Explanation / Answer
a) F=k(q1)(q2) / d^2 (k=9 x 10^9, q=charge of particle, d=distance between particles) First due to q1, F=(9 x 10^9)(6.54 x 10^-9)(5.35 x 10^-9) / (.3)^2 F = 3.4989 x 10^-6 N to the left since repulsive force due to q3, F=(9 x 10^9)(5.35 x 10^-9)(3.24 x 10^-9) / (.1)^2 F = 1.56 x 10^-5 N downwards because attractive force To find magnitude, use pythagorean theorem (3.4989 x 10^-6)^2 + (1.56 x 10^-5)^2 = x^2 x = 1.5988 x 10^-5 N b) To find direction, use trigonometry tan x = 1.56 x 10^-5 / 3.4989 x 10^-6 x = 77.36 degrees from the negative x axis 180 + 77.36 = 257.36 degrees from positive x axis
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