Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5.33m. A third charge of 1.0 nC is placed 1.9 m away from q1. If the net electric force on this third charge is zero, what is q2? (Give your answer in nC

-7.89 * 10^ (-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^ (-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^ (-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F (q1, q3) + F (q2, q3)

F = (k*q1*q3) / 1.9² + (k*q2*q3) / 3.43²

F = (9 * 10^ (9) * 2.42 * 10^ (-9) * 1 * 10^ (-9)) / 3.61 + (9 * 10^ (9) * q2 * 1 * 10^ (-9)) / 11.7649

F = 6.033 * 10^ (-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^ (-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^ (-9)

=> q2 = - [6.033 * 10^ (-9) ]/0.765

q2 = - 7.89 * 10^ (-9) C