Three identical point charges of magnitude q are located on the vertices of an i

Question

Three identical point charges of magnitude q are located on the vertices of an isosceles triangle with its altitude oriented vertically. The altitude of the triangle is 4 cm and the base is 6 cm long. The resultant electrical force on the charge located at the upper vertex of the triangle has a magnitude of 0.500 N. What is q? I determined the length of the leg of the triangle to be 5 cm, the x components of each force cancel out because they have the same magnitude and are in opposite directions, drawing a line down the center of the isosceles triangle to form 2 right triangles I found the angles to be 90 degrees, 53.13 degrees, and 36.87 degrees, Heres my question since the resultant vector is 0.5 N I know that either the y component or the resultant of force 3 on 2 is =0.25 N since thats half of 0.5 N, what I dont know is wether the 0.25 is the y component or the resultant force. The value for q that I found assuming that 0.25 N is the y component is 2.94 x10^-7 C. I know this is a wordy problem but I appreciate your help, thanks!

Explanation / Answer

Since you have already identified the angles, I am simply finding Fnet.

F(net) = 2F* cos 36.9
0.5 = 2 * (9*10^9 * q^2 ) / (0.05)^2 * cos 36.9

Solve for q.

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