Two point charges are placed on the x -axis as follows: charge q 1 = 4.04 nC is

ID: 779329 • Letter: T

Question

Two point charges are placed on the x-axis as follows: charge q1 = 4.04 nC is located at x= 0.198 m , and charge q2 = 4.99 nC is at x= -0.297 m .

a)What is the magnitude of the total force exerted by these two charges on a negative point charge q3 = -6.00 nC that is placed at the origin?

b)What is the direction of the total force exerted by these two charges on a negative point charge q3 = -6.00 nC that is placed at the origin?

4)the force is zero

1)to the +x direction 2)to the -x direction 3)perpendicular to the x-axis

4)the force is zero

We know that force on a point charge kept in electric field is

F=Eq

Where E is the magnitude of electric field and q is the magnitude of charge.

So first we will find the electric field at origin.

So net field at origin will be ,

E=(Kq2/r2 2)-(kq1/r1 2) in the positive x axis

E=9[(4.99/0.2972)-(4.04/0.1982)]

E=418.32N/C in the negative x direction

So the force will be

F=-6*10-9*418.324

F=2.51*10-6N

And the direction will be positive x axis

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