9. [-/2 Points] DETAILS Example 19.1 Where Is the Net Force Zero? Three point charges lie along the x axis as shown in the figure. The positive charge q1 = 21.0 AC is at x = 2.00 m, the positive charge 92 = 6.50 uC is at the 2.00 m origin, and the net force acting on q; is zero. What is the x coordinate of 73. -2.00 - x- SOLVE IT + + 42 F23 93 F18 41 Conceptualize Because q: is near two other charges, it Three point charges are placed along the x axis. If experiences two electric forces. The forces lie along the the resultant force acting on q, is zero, the force same line in this problem as indicated in the figure. F 13 exerted by q, on 43 must be equal in Because q3 is negative while q, and 92 are positive, the magnitude and opposite in direction to the force F 23 exerted by 42 on 43- forces F 13 and F 23 are both attractive. Categorize Because the net force on 93 is zero, we model the point charge as a particle in equilibrium. Write an expression for the net force on F, = F 23 + F13 charge q; when it is in equilibrium: 191 1 1931 F, = -K 192193 1 + K=(2.00 - x)" si = 0 x2 Move the second term to the right side of K - - 1921 193 - = ke- 191/1931 the equation and set the coefficients of the x2 (2.00 - X)2 unit vector i equal: Eliminate K, and |93l and rearrange the (2.00 - x)2 1921 = x21911 equation: (4.00 - 4.00x + x2) (6.50 x 10-6 C) = x2(2.10 x 10-5 c) Reduce the quadratic equation to a simpler 14.5x2 + 26.0x - 26.0 = 0 form: Solve the quadratic equation for the X = m positive root: Finalize The second root to the quadratic equation is x = -2.51 m. That is another location where the magnitudes of the forces on q; are equal, but both forces are in the same direction, so they do not cancel. MASTER IT HINTS: GETTING STARTED | I'M STUCK! Consider three charged particles placed along the x-axis. The particle with q, = 9.53 nC is at x = 6.44 m and 92 = 4.70 nC is at x = 2.43 m. Where must a positive charge, q3, be placed such that the resultant force on it is zero? m