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ID: 1417044 • Letter: P

Question

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Three point charges lie on the x axis. Charge 1 (+9.3 mu C) is at the origin, charge 2 (-5.5 muC) is at x = 12 cm. and charge 3 (+4.5 muC) is at x = 15 cm. k = 8.99 times 10^9 N middot m^2/C^2. What is the direction of the total force exerted on charge 3? in positive x-direction in negative x-direction the total force is zero What is the magnitude of the total force exerted on charge 3? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

by coulAMB LAW

F=KQaQb/d2

so force on q3

F =8.99*109[{ 9.3*10-6*4.5*10-6/0.152} +{-5.5*10-6*4.5*10-6/0.032} ]

F= 8.99*10-3[{ 9.3*4.5*/0.152} +{-5.5*4.5/0.032} ]

F= 8.99*10-3[1860-27500] = -230.5036N