Two negative equally charged drops water 380 mm apart exert

Two negative, equally charged drops water, 3.80 mm apart, exert an electrostatic force of 2.8 Times 10^-9 N on each other. Determine the net charge on each drop. 4.34 Times 10^(-2) nC. 6.7 Times 10^(-12)C. 7.69 Times 10^(3) pC. 7.17 Times 10^(-3) pC 8.19 Times 10^(3) pC. Then determine the number of electron that had been added to each drop to cause the net charge on it. 1.28 Times 10^(8) 3.73 Times 10^(9). 2.26 Times 10^(3). 9.26 Times 10^(6) 4.22 Times 10^2(7).

Solution

force F = Kq1q2/r^2

here F = 28.5 *10^-12 N (THIS IS NOT CLEAR)

r = 3.8 mm

q1 = q2 = q

so

q^2 = Fr^2/k

solve for q using correct F

(here we get q)

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use q = ne

where e = fundamental charge = 1.6 *10^-19C

so

n = q/e