Exercise 15.2 Hints: Getting Started | I'm Stuck Three charges lie along the x-axis. This is nearly always the case when quadratic equations are involved. m Remarks Notice that it was necessary to use physical reasoning to choose between the two possible answers for X. Put this equation into standard quadratic form, ax2 + bx + c = 0. (16 x 10-5C) | (6 m - 2) (10 x 10-6 C) les! ke = 0 Cancel kel 10-6 and 93 from the equation, concordant 1066 x)2 = 16x2ฤก0(6 - x)2 = 16x2 Cancel Iker 10-6 and 43 from the equation, and rearrange terms (explicit significant figures and units are temporarily suspended for clarity). Solution Write the x-component of F 13 Fix - the (16 x 10-6 c)|es (6 m -.-) (10 x 10-C) los Write the x-component of F23 F23 = - Set the sum equal to zero. The solution can be obtained with the quadratic formula. Write F 13 and F23 in terms of the unknown coordinate position x, sum them and set them equal to zero, solving for the unknown. Consequently, 93 must le between the other two charges. If the resultant force on 93 is zero, then the force F13 exerted by 91 on 43 must be equal in magnitude and opposite the force F23 exerted by 92 on 93 Strategy If q3 is to the right or left of the other two charges, then the net force on q3 can't be zero, because then F 19 and F 23 act in the same direction. The charge 43 is negative, whereas 91 and 42 are positive. Where must a negative charge q3 be placed on the x-axis so that the resultant electric force on it is zero? 2.0 m- 92 91 F9g 9 FS Figure 15.7 Three point charges are placed along the x-axis. The positive charge 41 = 16 PC is at x = 6 m, and the positive charge 42 = 10 pc is at the origin. Problem Three charges lie along the x-axis as in Figure 15.7. MY NOTES ASK YOUR 2.0 m Example 15.2 May the Force Be Zero Goal Apply Coulomb's law in one dimension.
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