Recent content by Anthony Santelices

I've tried it with the negative signs, but I end up getting -68. The correct answer shown was -105, so I don't know exactly where I went wrong.

These are the numbers I got along with the added ##q_1, q_3## $$ U_{c,q1} = \frac {(9*10^9)(+2.0*10^{-5}) (+10.0*10^{-5})} {2.0} = 9\ J$$ $$ U_{c,q2} = \frac {(9*10^9)(+2.0*10^{-5}) (-20.0*10^{-5})} {2.0} = 18\ J$$ $$ U_{c,q3} = \frac {(9*10^9)(+2.0*10^{-5}) (+10.0*10^{-5})} {2.0} = 9\ J$$ $$...

Ah yes! It appears I did. However, even without having it I was getting 126 J. It says that the answer is -105 J. So that tells me I messed up somewhere while calculating with all the ##U_e##'s. I don't have access to my work atm as I am not home. I will make sure to post it as soon I get the...

Homework Statement An object with charge +2.0×10−5 C is moved from position C to position D in the figure (Figure 1) . q1 = q3 = +10.0×10−5 C and q2 = −20.0×10−5 C. All four charged objects are the system. Here's a picture to the problem Homework Equations $$ F = \frac {kq_1 q_2}{r^2} $$ $$...