1. Two electric charges placed a certain distance away from each other feel an e

Question

1. Two electric charges placed a certain distance away from each other feel an electrical force (the "old force".) The charges are then moved apart so that they are five times the original distance from each other. At this new distane they feel a different force (the "new force".) What is the ratio of the new force to the old force?

2, A +3.0x10^-5 C charge and an +8.0x10^-4 C charge are placed 3 meters apart. Given that the constant in Coulomb's law is k = 9.0 x 10^9 Nm^2/C^2, what is the force (on either) charge?

Explanation / Answer

1) old Force is F_old = k*q1*q2/r^2

here k is proportionality constant

q1 and q2 are the charges

r is the distance if seperation

Given that new distacne is 5r

then new force F_new = k*q1*q2/(5*r)^2 = k*q1*q2/(25*r^2)
then required ration is

F_new/F_oid = [k*q1*q2/r^2] / [k*q1*q2/25*r^2] = (1/25)

F_new/F_old = 1/25

2) Force on either charge F = k*q1*q2/r^2

k = 9*10^9 Nm^2/C^2

q1 = 3*10^-5 C

q2 = 8*10^-4 C

r = 3m

then F = k*q1*q2/r^2 = (9*10^9*3*10^-5*8*10^-4)/3^2 = 24 N

repulsive force

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