Hey i dont know how to solve this. Can somebody help me?

In the two-plate arrangement shown in the sketch, there is homogeneous electrical field strength $$\vec{E_0} = -\vec{e}_y * 10\frac{kV}{cm}$$

enter image description here

Calculate the work done by the field when the charge Q = 1 μC from Point P 1 (l, a, 0) is moved to point P 2 (0, 0, 0). Integrate once over for practice the direct, sloping path and once over the square path along the Coordinate axes.

This my try to solve this task ... But i dont know if it is right $$ r = (-l\vec{e}_x -a\vec{e}_y) * t $$ $$ 0 <= t <= 1$$ $$ dr = (-l\vec{e}_x -a\vec{e}_y) * dt $$

$$ W = Q * \int_{0}^{1}\vec{E} dr$$


$$ W = Q* E_0 * \int_{0}^{1} -\vec{e}_y* (-l\vec{e}_x - a\vec{e}_y)dt$$ $$ W = Q* E_0 * -\vec{e}y* (-l\vec{e}_x - a\vec{e}_y)$$ $$ W = Q* E_0 * a$$

Thanks to the answers. I think this is the solution.

  • 1
    \$\begingroup\$ Please post your answer as an answer, not an update to the question, thanks \$\endgroup\$
    – Voltage Spike
    Dec 27 '17 at 19:14

Intuition: Work done by the field is Charge multiplied by the Potential Difference between the points. $$ W =Q.dV$$ Since a uniform vertical field is assumed, path taken by the charge between the plates is irrelevant. If the perpendicular distance is a, $$dV = E_o .a$$ $$ ie, W = QE_o a$$

  • \$\begingroup\$ Charge path is irrelevant in any case \$W=Q\Delta V\$ ,no matter the field being uniform. The latter just helps working out \$\Delta V\$ \$\endgroup\$
    – carloc
    Dec 27 '17 at 22:56
  • \$\begingroup\$ Yes I mean in calculating dV from E :-) \$\endgroup\$ Dec 28 '17 at 12:56
  • \$\begingroup\$ Yes I am sure that's what you mean but I read "Since a uniform vertical field is assumed, path taken by the charge between the plates is irrelevant." I am not native English speaker but it sounds different to me. :) \$\endgroup\$
    – carloc
    Dec 28 '17 at 13:16

Looks about right. Next step is to do the multiplication between \$\vec{E}\$ and \$d\vec{r}\$ keeping in mind that \$d \vec{r} \$ is also a vector. So this needs to be a vector product.

Sorry for not writing the symbols properly. Apparently this site doesn't support latex math tags.

  • 1
    \$\begingroup\$ This website does not have LaTeX. It has Mathjax which is similar to LaTeX. So if you want to use these symbols, your syntax has to be \$ this \$ and it'll look like \$this\$. Otherwise, your commands for characters are the same for LaTeX syntax. \$\endgroup\$
    – user103380
    Dec 27 '17 at 17:54
  • 1
    \$\begingroup\$ This might help. \$\endgroup\$ Dec 27 '17 at 18:23
  • \$\begingroup\$ i use double dollars. When you click edit you can look how i did this $$"xxx"$$ \$\endgroup\$
    – otto
    Dec 27 '17 at 18:30
  • \$\begingroup\$ @RohatKılıç The Math SE has a slightly different way of initializing equations for Mathjax. On their website, it's $ equation $ whereas, here, it's \$ equation \$. I don't know why it's different. \$\endgroup\$
    – user103380
    Dec 27 '17 at 19:21
  • 1
    \$\begingroup\$ apparently engineers write costs often which means they use the $ sign frequently, so you can't use a single $ to trigger mathjax, it's escaped to \$ \$\endgroup\$
    – Neil_UK
    Dec 27 '17 at 19:58

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