Three charged particles are placed at the corners of an equilateral triangle of

Question

Three charged particles are placed at the corners of an equilateral triangle of side d = 1.00 m (Fig. 16-53). The charges are Q1 = +4.0 µC, Q2 = -6.0 µC, and Q3 = -6.0 µC. Calculate the magnitude and direction of the net force on each due to the other two.

Force on Q1: ----N at----counterclockwise from +x axis (to the right)
Force on Q2: ----N at----counterclockwise from +x axis (to the right)
Force on Q3: --- N at--- counterclockwise from +x axis (to the right) Please help me. I am unable to do it. thanls
Three charged particles are placed at the corners of an equilateral triangle of side d = 1.00 m (Fig. 16-53). The charges are Q1 = +4.0 C, Q2 = -6.0 C, and Q3 = -6.0 C. Calculate the magnitude and direction of the net force on each due to the other two. Force on Q1: ----N at----counterclockwise from +x axis (to the right) Force on Q2: ----N at----counterclockwise from +x axis (to the right) Force on Q3: --- N at--- counterclockwise from +x axis (to the right) Please help me. I am unable to do it. thanls

Explanation / Answer

Hi, Force on Q1:     Force due to charge Q2 = F12 = k[(4 * 6)] * 10^(-12) / d^2                                      = 9 * 10^9 * [ 24 ] / 1 = 0.216 N towards Q2.     Force due to charge Q3 = F13 = k[(4 * 6)] * 10^(-12) / d^2                                      = 9 * 10^9 * [ 24 ] / 1 = 0.216 N towards Q3.                                      = 9 * 10^9 * [ 24 ] / 1 = 0.216 N towards Q3.     The resultant force on Q1 will be the vector sum of the above two forces. Resolving the two forces into perpendicular components, and observing the fact that the two forces are equal in magnitude and acting symmetrically on opposite sides of the bisector of the line joining Q2 and Q3, the x-components of the two forces will be opposite to each other and equal in magnitude and so will get cancelled.     Hence the resultant force on Q1 = (F12 * Sin30) + (F13 * Sin30)                                                    = 0.216 N     and in the direction -ve y axis i.e., at an angle of 270degrees counter clockwise from +ve X-axis. For charge Q2:     Force due to charge Q1 = F21 = k[(4 * 6)] * 10^(-12) / d^2                                      = 9 * 10^9 * [ 24 ] / 1 = 0.216 N towards Q1 (60 degrees with +ve x-axis and 30 degrees with +ve y-axis as seen from the geometry of triangle).     Force due to charge Q3 = F23 = k[(6 * 6)] * 10^(-12) / d^2                                      = 9 * 10^9 * [ 36 ] / 1 = 0.324 N away from Q3 (-ve x-axis.)                                      = 9 * 10^9 * [ 36 ] / 1 = 0.324 N away from Q3 (-ve x-axis.)         Hence the x-component of the resultant force on Q2       = F2x = F21x + F23x = (0.216 * Cos60) + (0.324 * Cos180)                                   = 0.108 - 0.324 = -0.216 N     similarly the y-component on Q2       = F2y = F21y + F23y = (0.216 * Sin30) + (0.324 * Sin180)                                   = 0.108 - 0 = 0.108 N     Hence the resultant force on Q2 = sqrt(F2x^2 + F2y^2) = sqrt(0.046656 + 0.001164)                                          = 0.242 N in the direction given by arctan(F2y/F2x) . Follow the similar approach to get the values or Q3. Hope this helps you. Follow the similar approach to get the values or Q3. Hope this helps you.

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