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)

Answer 1

Hi,Force on Q1:Forcedue to charge Q2 = F12 = k[(4* 6)] * 10^(-12) / d^2=9 * 10^9 * [24]/ 1 = 0.216N towards Q2.Force due to charge Q3 = F13 = k[(4* 6)] * 10^(-12) / d^2=9 * 10^9 * [24]/ 1 = 0.216N towards Q3.The resultant force on Q1 will be the vector sum of the above two forces. Resolving the two forces into perpendicularcomponents, and observing the fact that the two forces are equal in magnitude and acting symmetrically on oppositesides of the bisector of theline 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)=0.216N and in the direction-vey axis i.e., at an angle of 270degrees counter clockwise from +ve X-axis.For charge Q2:Forcedue to charge Q1 = F21 = k[(4* 6)] * 10^(-12) / d^2=9 * 10^9 * [24]/ 1 = 0.216N towards Q1 (60 degrees with +ve x-axis and 30 degrees with +ve y-axis as seen from the geometry oftriangle).Force due to charge Q3 = F23 = k[(6* 6)] * 10^(-12) / d^2=9 * 10^9 * [36]/ 1 = 0.324Naway fromQ3 (-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 Nsimilarly the y-componenton Q2= F2y= F21y + F23y = (0.216 * Sin30) + (0.324 * Sin180)=0.108 - 0 = 0.108 NHence the resultant force on Q2 = sqrt(F2x^2 + F2y^2)= sqrt(0.046656 + 0.001164)=0.242 N in the direction given byarctan(F2y/F2x) .Follow the similar approach to get the values or Q3.Hope this helps you.