Asked by Traci
Two identical point charges are fixed to diagonally opposite corners of a square that is 1.5 m on a side. Each charge is +2.0 µC. How much work is done by the electric force as one of the charges moves to an empty corner?
I have W(fe)= -EPE=-q[V(f)-V(i)]. I can't figure out what to do next or if im doing it right to begin with?
I have W(fe)= -EPE=-q[V(f)-V(i)]. I can't figure out what to do next or if im doing it right to begin with?
Answers
Answered by
Anonymous
what next? Figure out the potential at the original corner, and the potential at the new corner
V=kQ/distance
the distance from the furthermost corner changes
V=kQ/distance
the distance from the furthermost corner changes
Answered by
Traci
I'm unsure of how to figure out the potential at the different corners because if the first distance is kq/1.5, why isnt the 2nd one kq/1.5?
Im doing (9e9)(2.00e-6)/1.5=12000, which I don't know which number that is, or what to do with it further?
Im doing (9e9)(2.00e-6)/1.5=12000, which I don't know which number that is, or what to do with it further?
Answered by
Derek UNCC
If your sides are equal, the original distance to the corner is found by the hypotenuese of the sides. Which is 2.1213m This is your original to the far corner, when moving to the side it comes down to 1.5. Plug those numbers in.
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