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Work Done line Integral question - Electrostatics - help please
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[QUOTE="BvU, post: 5098083, member: 499340"] You showed that ##\vec E## as given is curl free. Well done. Then you worked out that ##\displaystyle q\int_{(0,0,0)}^{(x_1, y_1,0)} \vec E\cdot \vec{dl} = q\left(3x_1^2y_1 - y_1^3\right )## (small omission of the ##^2## ). To check it, you could verify that ##V = -\nabla\cdot E## is satisfied if ##-V(x,y,z) = 3x^2y - y^3## and then you can immediately see that W.D. is what you found (when you fix the missing square) and not q(3x[SUP]2[/SUP]-6y). You could double check with another path: from (0,0,0) to (0, y[SUB]1[/SUB], 0) and then to ( x[SUB]1[/SUB], 0, 0) which givves the very same result. Have some faith and if we're both wrong don't hesitate to inform me with another post ! I love (:wink:) to be corrected when I'm wrong. (happens all the time :cry: ) - [/QUOTE]
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Introductory Physics Homework Help
Work Done line Integral question - Electrostatics - help please
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