Using the same triangle, find the vector components of the electric force on q1

Question

Using the same triangle, find the vector components of the electric force on q1 and the vector's magnitude and direction.(Use the charges given in the Practice It section.)

Fx=

Fy=

Magnitude=

Direction =_____degrees counterclockwise from the + X-axis

Previous: Consider three point charges at the corners of a triangle, as shown in the figure above, where q1 = 6.38 10-9 C, q2 = -1.54 10-9 C, and q3 = 5.11 10-9 C

Using the same triangle, find the vector components of the electric force on q1 and the vector's magnitude and direction.(Use the charges given in the Practice It section.) Fx= Fy= Magnitude= Direction =_____degrees counterclockwise from the + X-axis Previous: Consider three point charges at the corners of a triangle, as shown in the figure above, where q1 = 6.38 10-9 C, q2 = -1.54 10-9 C, and q3 = 5.11 10-9 C

Explanation / Answer

Look at the picture : Fa = F12 = kq1*q2/r12^2 = 9 x 10^9 * 6.44 10^ -9 * 1.59 10^ -9/3^2 = 1.02 x 10^-8 N Fb = F13 = kq1*q3/r13^2 = 9 x 10^9 * 6.44 10^ -9 * 5.15 10^ -8/3^2 = 1.19 x 10^-7N a) Fx = Fax + Fbx = 1.02 x 10^-8 cos 90 + 1.19 x 10^-7 cos 37 = 0 + 9.5 x 10^-8 = 9.5 x 10^-8 N b) Fy = Fay + Fby = 1.02 x 10^-8 sin 90 + 1.19 x 10^-7 sin 37 = 8.18 x 10^-8 N c) R = v(Fx^2 + Fy^2) = v(9.5 x 10^-8^2 + 8.18 x 10^-8^2) = 1.25 x 10^-7 N d) tan ? = Fy/Fx ? = arc tan (8.18 x 10^-8/ 9.5 x 10^-8) = 40.73 degrees

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