# Two point charges, A = -2Q and B =6Q, are located at the positions shown in the figure above....?

ok i dont have the picture but ill just explain it...Two point charges, A = -2Q and B =6Q, are located at the positions shown in the figure above,on the x,y plane the A=-2Q charge is placed on the left side of the y-axis,in the negative x region 3meters away from the origin,and B=6Q is 2 meters right above the origin on the positive side of the y-axis...

a) What is the force on a charge q at the origin?

the answer is in the form of ( i + j) kQq N

b) Where would you place a point charge +Q such that the net force on q is zero?

so for the first part i just tried to use coulombs law force=(K(Q)(q))/distance^2

and it doesnt seem to be right for some reason...doesn kQ and kq mean kilocoulomb?thats what i did in my equation...

((9*10^9) N*m^2*c^2 * (-2*10^-3)kQ * (1*10^-3)kq)/ 3^2 for the origin and A

((9*10^9) N*m^2*c^2 * (6*10^-3)kQ * (1*10^-3)kq)/ 2^2 for the origin and B

my answer i get (-2000 i+13500 j) kQq N which is wrong...what am i doing wrong?

### 1 Answer

- kuiperbelt2003Lv 79 years agoFavorite Answer
k Q does not mean kilo Coulomb (that would be a huge charge), but comes from Coulomb's Law

F= k q1 q1/r^2 where k is a constant, q1, q2 are the charges, and r is the separation

in this case, calculate the force along the x axis due to the charge at (-3,0), and then the force along the y axis from the charge at (0,2)

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