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Moneeka96
Pleaseeee help me.. What is the electric force between a +2 µC point charge and a –2 µC point charge if they are separated by a distance of 5.0 cm? (µC = 1.0 × 10–6 C)
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Coulomb's Law: \[F=K{q_1q_2\over d^2}\] \[K = const.\\ q_1 = 2\times10^{-6}C\\ q_2= 2\times10^{-6}C\\ d=5{\rm cm}\times {1{\rm m}\over 100{\rm cm}}=0.05{\rm m} \]
the important thing is to have all the units in SI system
So i had to put cm into m like i believe you did above?
now from here where do I go?
wouldnt q2 be negative?
we plug em in the equation for the force. \[ F=8.99\times10^{9}\times \frac{2\times10^{-6}\times2\times10^{-6}}{(0.05)^2} \]
the negative sign can be neglected since we are only looking for the magnitude (i.e., the amount) of force. since the two charges are oppositely charged, the above force will be a force of attraction. this is the information we get from the charges
Okay cool so then Id just plug the above into my calculator and thats the answer. right?