Two point charges are placed on the x axis as follows: Charge q 1=+3.90nC is loc

Question

Two point charges are placed on the x axis as follows: Charge q1=+3.90nC  is located at x=0.225m , and charge q2=?4.75nC is at x=+0.405m . What is the magnitude of the net force exerted by these two charges on a negative point charge q3=?0.625nC placed at the origin?

What is the direction of the net force exerted by these two charges on a negative point charge q3=?0.625nC placed at the origin?

** Show Work Please

Problem 17.22 Part A Two point charges are placed on the x axis as follows: Charge q! = +3.90nC is located at x = 0.225m , and charge q2-_4.75nC is at x = +0.405m . What is the magnitude of the net force exerted by these two charges on a negative point charge q3--0.625nC placed at the origin? Express your answer in newtons to three significant figures. Fnet = Submit My Answers Give Up Part B What is the direction of the net force exerted by these two charges on a negative point charge q3--0.625nC placed at the origin? O - direction direction Submit My Answers Give Up

Explanation / Answer

PART A

Negative charges are attracted to positive charges. The effect of a charge decreases with the square of the distance between the charges.
Coulombs Law says F=k*Q1*Q2/r^2 where k = 9*10^9 N*m^2/C^2,
Q1 & Q2 are the charges in Coulombs and r is the distance between them in Meters.

The problem requires you to calculate the force of q1 on q3 and the force of q2 on q3 separately and then add them to get the net force.

I'm assuming that nC is a nano-Coulomb (1/10^9).
F1 = 9*3.90*0.6/(.225)^2 = 416 N in the +X direction
F2 = 9*4.75*.6/(.405)^2 = 156.3 N in the -X direction

PART B

The net force is F1+F2 = 572.3 N in the +X direction.

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