Three point charges, +6.7 µC, +2.5 µC, and 3.0 µC, lie along the x-axis at 0 cm,

Question

Three point charges, +6.7 µC, +2.5 µC, and 3.0 µC, lie along the x-axis at 0 cm, 2.4 cm, and 5.7 cm, respectively. What is the force exerted on q1 by the other two charges? (To the right is positive.) The Coulomb constant is 8.99 × 109 N · m2 /C 2 . Answer in units of N.

part b

What is the force exerted on q2 by the other two charges? (To the right is positive.) Answer in units of N.

part c

What is the force exerted on q3 by the other two charges? (To the right is positive.) Answer in units of N.

Explanation / Answer

k =8.99x10^9 N.m^/C2

q1 =6.7 µC, q2 =2.5 µC q3 = -3 µC

x1 ==0cm , x2 =2.4 cm , x3 =5.7 cm

F =kQ1Q2/r^2

Part a:

F1 = F12 +F13

F12 = kq1q2/r^2 = (9x109x6.7x10-6x2.5x10-6)/(0.024*0.024)

F12 = - 261.72 N (repulsive force)

F13 = kq1q3/r^2 = (9x109x6.7x10-6x3x10-6)/(0.057*0.057)

F13 = +55.68 N (attractive force)

F1 = -261.72+55.68 = 206.04 N

(b) F21= kq1q2/r^2 = (9x109x6.7x10-6x2.5x10-6)/(0.024*0.024)

F21 = 261.72 N (repulsive force)

F23= kq3q2/r^2 = (9x109x3x10-6x2.5x10-6)/(0.057*0.057)

F23 = 20.78 N (attractive force)

F2 =F21 +F23 = 261.72 +20.77 N

F2 =282.5 N

Part (c)

F31 = kq1q3/r^2 = (9x109x6.7x10-6x3x10-6)/(0.057*0.057)

F31 = -55.68 N (attractive force)

F23= kq3q2/r^2 = (9x109x3x10-6x2.5x10-6)/(0.057*0.057)

F23 = - 20.78 N (attractive force)

F3 =F31+F23 = -55.68-20.78

F3 = 70.46 N

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