Three point charges are arranged along the x-axls. charge q1 = +4.25 μc is at the origin, and charge q2 =-5.75 pc ls at x = 0.230 m. charge q3 =-9.00 pc. where ls σ3 located if the net force on q1 ls 6.00 n in the -x-direction?

Hey there!:

q1 = 4.25 * 10^-6 C is at x1 = 0 m

q2 = - 5.75 * 10^-6 C is at x2 = 0.23 m

q3 = - 9 uC = -9 * 10^-6 C

for the net force on q1 to be in -x direction

q3 must be placed to the left of q1

the net force on q1 , F = k * q1 * q3 /( x3^2) - k * q1 * q2 /( x2 - x3)^2

6 = 9 * 10^9 * 10^-6 * 4.25 *  10^-6 * (9 /x3^2 - 5.75 /( 0.25 - x3 )^2 )

solving for x3

x3 = - 0.22 m

the charge q3 is placed at x3 =  - 0.22 m

Hope this helps!

i wanna say it’s between b and d, leaning more towards b.