Three charges are at the corners of an isosceles triangle as shown in the figure

Question

Three charges are at the corners of an isosceles triangle as shown in the figure (Figure 1) . The +q1 = 5.25 *10^6 C and -q1= 5.25 * 10^6 C charges form a dipole. a) Find the magnitude and direction of the net force that the q2= 9.50 *10^6 C charge exerts on the dipole. b) Is the force upward or downward? c) For an axis perpendicular to the line connecting the two charges of the dipole at its midpoint and perpendicular to the plane of the paper, find the magnitude and direction of the torque exerted on the dipole by the q2= 9.50 *10^6 C charge. d) Is the torque clockwise or counterclockwise? If possible, a brief explanation of how you approach each portion of the problem would be very helpful. I've tried to solve this problem, but my professor worked the problem so fast that I did not get a firm grasp of the approach. Thank you in advance!

Explanation / Answer

PLEASE RATE ME AND AWARD ME KARMA POINTS IF IT IS HELPFUL FOR YOU The force on q1 by q2 acts at angle 23 degree from the x-axis, and the other force act an angle 157 degree from the x-axis. Since their magnitudes are equal, the horizontal component of the forces are obtained from cos 23 and cos 157 which are equal but opposite. Therefore they cancel each other. No net force is acting in the horizontal direction. The vertical components are found using sin 23 and sin 157. sin23 = sin 157.Hence the vertical force is two times any one of the forces. The net force = 2.K.(sin 23).(9 x 5.5) / (1.3)^2= 22.9 K. The value of K = 9 x10^9 µ µ = 9 x10^ (-3). The force = 22.9*{9 x10^ (-3)} = 0.206 N. b) Mass =1.2 x 10^(-3) kg. Acceleration is force / mass = 0.206 N/ 1.2 x 10^(-3) = 172m/s^2.

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