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?

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

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min