Hi all. I don't know if I'm completely over thinking this question or just not getting it. Please tell me if I'm correct. A -8.0μC charge is located 0.30m to the left of a +6.0μC charge. What is the magnitude and direction of the electrostatic force on the positive charge? F=KQaQb/r^2 F=(9x10^9)(-8.0x10^-6)(+6.0x10^-6)/(0.30m)^2 = -4.8N...so answer is 4.8, to the right. So, I think that because the F is -4.8 means that the direction is going to the right because a negative means that it is an attractive force and the +6 charge will be attracted to the -8 charge. Am I correct?