Physics help plzzzzzzzzzzzzz?

Question

A charge q1 of -5*10 ^ -9 and the charge q2 of -2 * 10 ^-9 C are seperated by the distance of 40 cm. Find the equilibrium position for a third charge +15*10 ^ -9 C.

Answer 1

(-5)(15)/ x^2 =(-2)(15)/ y^2
.
-5/x^2=-2/y^2
5 y^2= 2x^2
.x=y(sqrt2.5) and x+y=40
x=~1.51 y and 1.51y+y=40 ,,2.51y=40,y=40/2.51cm
y=~16cm x=~24cm

Answer 2

let the third charge be q3. Equillibrium occurs when the forces due to q1 and 2q on q3 are equal and opposite in direction, hence q3 is between the two negative charges. Let d = 40 cm.

The total force F = 0 = k*q1*q3/x^2 -k*q2*q3/(d-x)^2

where x = distance from charge 1. Can divide out a k*q3

0 = q1/x^2 - q2/(d-x)^2 --- > q1/x ^2= q2/(d-x)^2

x^2/q1 = (d-x)^2/q2 ---> x^2 = (q1/q2)*(d-x^2)

x^2 = (q1/q2)*(x^2 - 2xd +d^2)

0 = (q1/q2-1)*x^2 -2(q1/q2)xd +q1/q2 * d^2

0 = x^2 -2axd+ ad^2 a = q1/q2/(q1/q2-1)


Then

x = ad +/- sqrt((2ad)^2 - 4ad^2)/2 = ad +/- d*sqrt(a^2-a)

I leave the numerial substitution up to you. Just remember that x < 40 cm.