Question

Coulomb's Law: As shown in the figure, charge q1 = 2.2 × 10-6 C is placed at the origin and charge q 2 = -3.30 × 10-6 C is placed on the x-axis, at x = -0.200 m. Where along the x-axis can a third charge Q = -8.30 × 10-6 C be placed so that the resultant force on Q is zero?

Answer: .89 m

Please show work on how to get this answer

Answer 1

The resultant force on the third charge will be zero when the electric fields E from q1 and q2 cancel leaving an E field intensity of magnitude zero.

Now, since q1 and q2 are opposite polarity, their fields add between them since they are in the same direction there. To the left or right, the fields subtract. In my opinion, the point will be to the right
since 2.2 < 3.3
E = kq1/r1^2 - kq2/r2^2 = 0
r1 = d where d is the distance to the right of q1
r2 =(0.2+d)

Now, factor out the k and 10^-6 because they appear in both expressions.

So, the resulting expression -

2.2/d^2 = 3.3/(0.2+d)^2 This makes sense because we know the fields are in opposite directions to the right of q1. q1 points to the right and q2 to the left. Thus when they are equal, they cancel exactly.

So -

2.2(.2+d)^2 = 3.3d^2
=> 1.1d^2 -0.88d -0.088 = 0
Solve this quadratic equation, we get -

d = 0.8898979 = 0.89 m.

So, the position of the third charge on the x - axis is 0.89 m from the origin.