# Electric Field electrical charge

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 $$\vec{E_0} = -\vec{e}_y * 10\frac{kV}{cm}$$ 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 $$r = (-l\vec{e}_x -a\vec{e}_y) * t$$ $$0 <= t <= 1$$ $$dr = (-l\vec{e}_x -a\vec{e}_y) * dt$$

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

UPDATE1

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

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

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, $$dV = E_o .a$$ $$ie, W = QE_o a$$

• Charge path is irrelevant in any case $W=Q\Delta V$ ,no matter the field being uniform. The latter just helps working out $\Delta V$ Dec 27 '17 at 22:56
• Yes I mean in calculating dV from E :-) Dec 28 '17 at 12:56
• 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. :) 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.

• 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.
– user103380
Dec 27 '17 at 17:54
• This might help. Dec 27 '17 at 18:23
• i use double dollars. When you click edit you can look how i did this $$"xxx"$$
– otto
Dec 27 '17 at 18:30
• @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.
– user103380
Dec 27 '17 at 19:21
• 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 \\$ Dec 27 '17 at 19:58