In the figure below, charge q1 = 2.2 x 10-6 C is placed at the origin and charge

Question

In the figure below, charge q1 = 2.2 x 10-6 C is placed at the origin and charge 1)
q2 = -3.3 x 10-6 C is placed on the x-axis, at x = -0.20 m.

Where along the x-axis can a third charge Q = -8.3 x 10-6 C be placed such that the
resultant force on this third charge is zero?

Please explain clearly, this is always confusing for me. I know since the Q is strongly negative it goes to the right of the positive q1.

In the figure below, charge q1 = 2.2 x 10-6 C is placed at the origin and charge 1) q2 = -3.3 x 10-6 C is placed on the x-axis, at x = -0.20 m. Where along the x-axis can a third charge Q = -8.3 x 10-6 C be placed such that the resultant force on this third charge is zero? Please explain clearly, this is always confusing for me. I know since the Q is strongly negative it goes to the right of the positive q1.

Explanation / Answer

Net force on the third charge Q must be 0.

F= Kq1q2/r^2

let the third charge is placed at r distance riight from q 1

r13 = r

r23= r+0.2

F13 + F23 = 0

Kq1q3/(r13)^2 + Kq2q3/(r23)^2 = 0

=> q1q3/(r13)^2 + q2q3/(r23)^2 = 0

(2.2 x 10-6 C )* ( -8.3 x 10-6 C )/ r^2 + (-3.3 x 10-6 C )*( -8.3 x 10-6 C )/(r+0.2)^2 = 0

-18.26(r+0.2)^2 * 10^-12 + 27.39 (r^2) * 10^-12 = 0

-18.26 r^2 - 7.304 r - 0.7304 + 27.39r^2 = 0

9.13 r^2 - 7.304 r - 0.7304 = 0

solving the quadratic equation

r1= (7.304 + sqrt(7.304^2 - 4 * 9.13 * (-0.7304))) / 2*9.13

r1= (7.304 + sqrt (80.02264))/18.26

r1=0.89 m

r1= (7.304 - sqrt(7.304^2 - 4 * 9.13 * (-0.7304))) / 2*9.13

r1= (7.304 - sqrt (80.02264))/18.26

r1=-0.090 m

The charge can be placed 0.89 m right of q1 or 0.090 m left of q1 so that net force is 0.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---