Goal Apply Coulomb\'s law in one dimension. Problem Three charges lie along the

Question

Goal Apply Coulomb's law in one dimension.

Problem Three charges lie along the x-axis as in Figure 15.7. The positive charge q1 = 17 µC is at x = 7 m, and the positive charge q2 = 6 µC is at the origin. Where must a negative charge q3 be placed on the x-axis so that the resultant electric force on it is zero?
Figure 15.7 Three point charges are placed along the x-axis. The charge q3 is negative, whereas q1 and q2are positive. If the resultant force on q3 is zero, then the force 13 exerted by q1 on q3 must be equal in magnitude and opposite the force 23 exerted by q2on q3. 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 13 and 23 act in the same direction. Consequently, q3 must lie between the other two charges. Write 13 and 23 in terms of the unknown coordinate position x, sum them and set them equal to zero, solving for the unknown. The solution can be obtained with the quadratic formula. Solution Write the x-component of 13. Write the x-component of 23. Set the sum equal to zero. Cancel ke, 10-6 and q3 from the equation, and rearrange terms (explicit significant figures and units are temporarily suspended for clarity). 6(7 - x)2 = 17x2 Put this equation into standard quadratic form, ax2 + bx + c = 0. 6(49 - 14x + x2) = 17x2 294 - 84x + 6x2 = 17x2
11x2 + 84x - 294 = 0 Apply the quadratic formula. Only the positive root makes sense. x= 2.0 m 2.0 m.

Explanation / Answer

let the distance of charge Q3 from origin be "a"

F13 = force by charge 1 on charge 3 in right direction = k Q1 Q3/(7 - a)2

F23 = force by charge 2 on charge 3 in left direction = k Q2 Q3/(a)2

for the Net force to be zero

F23 = F13

k Q2 Q3/(a)2 = k Q1 Q3/(7 - a)2

Q2 /(a)2 = kQ1/(7 - a)2

6 /(a)2 = 17/(7 - a)2

a = 2.61 m

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.