I have been using the provided formula and converting to Coulombs from nC and ev

Question

I have been using the provided formula and converting to Coulombs from nC and everything but I am not getting the correct answer, I think I am doing something incorrectly. I am getting -0.00001211

Forces in a Three-Charge System Part A What is the net force exerted by these two charges on a third charge q3-46.5 nC placed between q1 Coulomb's law for the magnitude of the force F between two particles with charges @ and Q separated by a distance d is and q2 at z,--1.250 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures where K = = 8.854 × 10 permttivity of free space. Consider two point charges located on the x axis: one charge, q1-12.5 nC, is located at 4TE, and -12 C2/(N-m*) is the Force on q3 = Submit Hints My Answers Give Up Review Part =-1.745 m ; the second charge, q2 = 34.5 nC is at the origin (z = 0.0000) Incorrect; Try Again; 2 attempts remaining

Explanation / Answer

From the given problem we say that q1 is negative and to the left of q3 then the force will be to the left

and again the q2 is positive and to the right of q3 then the force will aso be o the left

Therefore we can add the forces together

F =-kq1q3/r132 -kq2q3/r232

= -9*109*-12.5*10-9*46.5*10-9/(1.745 -1.250)2 --9*109*34.5*10-9*46.5*10-9/(1.250)2

=-2.1345*10-5 -0.9240*10-5

=-3.0585*10-5N    ( towards the left being negative)

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