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 force of 1.00 N when the spheres are 2.20 m apart, what is the charge on the sphere with the smaller charge?

Answer 1

the charge on the sphere with the smallest charge = 1.41×10^-5 C

Step-by-step 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