Ask Question
8 August, 17:21

Two small, positively charged spheres have a combined charge of 5.20x10^-5 C. If each sphere is repelled from the other by an electrostatic force of 1.00 N when the spheres are 2.20 m apart, what is the charge on the sphere with the smaller charge?

+5
Answers (1)
  1. 8 August, 17:29
    0
    the charge on the sphere with the smallest charge = 1.41*10^-5 C

    Explanation:

    let q1 and q2 be the charges on the spheres, r be the distance between the spheres, k = 9*10^9 [N. (m^2) ]/C be the coulombs constant.

    the electrostatic force is given by:

    F = k*q1*q2 / (r^2)

    1.00 = [ (9*10^9) * (q1*q2) ]/[ (2.20) ^2]

    (q1*q2) = 5.34*10-10 C^2

    q1 = [5.34*10-10 C^2] / (q2)

    From the information we given that:

    q1 + q2 = 5.20*10^-5 C

    [5.34*10-10/q2 + q2 = 5.20*10^-5

    (q2) ^2 - 5.20*10^-5*q2 + 5.34*10-10 = 0

    Aplplying the quadratic formular to the equation above we get:

    q2 = 3.79*10^-5 C or q2 = 1.41*10^-5 C

    for q2 = 3.79*10^-5 C

    then q1 = 5.20*10^-5 - 3.79*10^-5 = 1.41*10^-5 C

    Therefore, the charge on sphere with the smallest charge is 1.41*10^-5 C.

    then m1 = 5.80 - 1.05 = 4.75 kg
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Two small, positively charged spheres have a combined charge of 5.20x10^-5 C. If each sphere is repelled from the other by an electrostatic ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers