I've calculated the magnitude force but don't know from here.?

Question

Two point charges q1 = (+) 1.5µC and q2 = (+) 8.5µC are separated by a distance of 10cm. At what point on the line joining the two charges should a charge q = (-) 1µC be placed so that the net force on q due to q1 and q2 is zero?

Answer 1

Let it be placed at distance x cm from q1 i.e. 10-x cm from q2.

q1 and q2 pull q in oppsite directions. The magnitudes of their forces should be equal for net force on q to be zero.

Force on q by q1 = force on q by q2 in magnitude

Or, q1*q/x^2 = q2*q/(10-x)^2

Divide by q.

q1/x^2 = q2/(10-x)^2

Or, x^2/q1 = (10-x)^2/q2

Or, x^2/1.5 = (10-x)^2/8.5

Or, 8.5x^2 = 1.5*(10-x)^2

Or, 8.5x^2 = 1.5(100-20x+x^2)

Or, 8.5x^2 = 150 - 30x + 1.5x^2

Or, 7x^2 + 30x - 150 = 0

x = {-30 ± √(30^2 + 4*7*150)}/(2*7)

= {-30 ± √(5100)}/14

x must be positive.

Therefore, x = {-30 + √(5100)}/14

= (-30 + 71.4)/14

= 2.96

Ans: 2.96 cm from q1

Note: The previous answerer has used distance where he should use distance square.

Answer 2

So you have three charges, q1 and q2, then q in the middle. For the force between them q1 and q2 to be zero, we need the force between q1 and q, to equal the force between q2 and q. We will call these forces F1 and F2 respectively.

I'm assuming you have the formula for force on static charges

F = k*C1*C2 / d

where k is the electric constant, C1 and C2 are the value of the two charges, and d is the distance in meters between them.

if F1 = F2,

k*q1*q/d = k*q2*q/(0.1-d)

the (0.1 - d) comes from the fact that there is 0.1 meters distance between q1 and q2, a distance of d between q1 and q, and a distance of (0.1 - d) between q2 and q

we can get rid of the k and q on both sides of the equation, and it becomes

q1/d = q2/(0.1-d)

q1*0.1 - q1*d = q2*d

q1*0.1 = d(q1+q2)

d = (q1*0.1)/(q1 + q2)

You can make the substitutions yourself because i don't have a calculator, but q is the distance d away from q1, and q is a distance of (0.1 - d) distance away from q2.

Hope that helps.