- #1
Dopefish1337
- 47
- 0
Homework Statement
http://img413.imageshack.us/img413/8307/prob01a.gif
Determine the work done on a 0.5 µC charge moved from A to B.
Homework Equations
delta V= -integral(E*dl), aka the area under the curve on an E-field graph from A to B equals the potential difference. (Sorry, don't know how to display math properly on these forums yet.)
(delta V)*(q)=Work
The Attempt at a Solution
I've attempted to aproximate the the area under that curve 9 times now (from A to B) using a series of triangles/rectangles, and have gotten several results for the area (delta V) that ranged from 12V(certainly to small) to 18V (certainly to big).
I've been largely ignoring the sign's of things throughout, as the field is clearly from a positive charge, and so moving a 0.5*10-6 C charge closer to it will surely result in a positive amount of work done in the end. Is there any error in this reasoning, such that I should only be trying negative results?
Is there a better way to find the work done beyond attempting to aproximate the area? It makes theoretical sense to me so it 'should' work, but the absence of particularly precise y-values on the graph make it difficult, and it seems unreasonable of them to want a very accurate answer (which I presume they do, based on my first 9 tries being rejected) considering they didn't give us the equation of the graph.
Final answers I've tried ranged from 6.00*10-6 J to 8.00*10-6 J, but I've foolishly neglected to write down exactly what all I've tried because I was confident that my method was sound and it'd have a large range of acceptable answers to account for the aproximation of the area, but apparently not.
Last edited by a moderator: