Two point charges are placed on the x -axis as follows: charge q 1 = 3.95 nC is

Question

Two point charges are placed on the x-axis as follows: charge q1 = 3.95 nC is located at x= 0.198 m , and charge q2 = 5.00 nC is at x= -0.295 m .

Part A: What is the magnitude of the total force exerted by these two charges on a negative point charge q3 = -6.05 nC that is placed at the origin?

Part B: What is the direction of the total force exerted by these two charges on a negative point charge q3 = -6.05 nC that is placed at the origin?

Explanation / Answer

Here,

q1 = 3.95 nC

part A) for the total force acting on the q3 placed at the origin

Net force = 9 *10^9 * 10^-9 * 6.05 *10^-9 * (3.95/0.198^2 - 5/0.295^2)

Net force = 2.35 *10^-6 N

the net force on the third charge is 2.35 *10^-6 N

part b)

as the net force is positive

the direction force is

to the +x direction

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