Work Done line Integral question - Electrostatics - help please

[smileandbehappy]

Can anyone please tell me where I am going wrong? I am getting the incorrect answer for the Word Done should be: WD = q(3x^2-6y) ...

Apologies for not changing it into the format on here - but for my revision I have pretty much done that myself.

 

[BvU]

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(3x²-6y).

You could double check with another path: from (0,0,0) to (0, y₁, 0) and then to ( x₁, 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 () to be corrected when I'm wrong.

(happens all the time )
-

 

[smileandbehappy]

Thanks - I guess I was just doubting myself! Really appreciate you helping me out like that. Sam.