Question

A –4.0-μC charge is located 0.45m to the left of a +6.0-μC charge. What is the
magnitude and direction (to the right or to the left) of the electrostatic force on the
positive charge?

Answer 1

to find the force between two point charges, Coulomb's Law is applied....
\[F = \frac{ 1 }{ 4\pi \epsilon } \times \frac{ Q _{1}Q _{2} }{ r ^{2} }\]

Answer 2

F=Force
\[\epsilon \]=permittivity of free space.
Q1 = charge of one of the particles
Q2=charge on the other particle
r=distance apart

Answer 3

Thanks a million! It's probably an easy question but I'm horribly bad at physics. I'll give it a shot but if you have time and if it would be no hassle to you I'd love to see how you would go about doing it :)

Answer 4

For particles in a vacuum, \[\frac{ 1 }{ 4\pi \epsilon }\] is estimated to be 9 x 10^9

Answer 5

since the two charges are opposite, they would attract each other .
therefore, the positive charge would move towards the negative charge..which means it would move to the ???
I think you should try that..

Answer 6

you are welcome!!

Answer 7

Ok so obviously the charge would move to the left but I'm still not quite sure if I got the right answer? Can you check and see if it's right please?
F= k•(q1•q2)/r^2=9•10^9•(4•10^-6•6•10^-6)/(0.45)^2=1.067N
What you think??

Answer 8

that is right! :)

Answer 9

YES!!!!!!!!! Thank you! I just joined this today. Didn't think I'd get much help out of it but You have restored my faith! Delighted! I am waaay too excited!

Answer 10

you are welcome!! I am happy you have restored your faith....and I am also happy to have helped you....