0
$\begingroup$

If I calculate the electric field between two conducting objects using $2D$ FEM i.e.,

enter image description here

Can I then simply integrate the electric field around the boundary of an object to get the net electric force acting on it? (because $F=qE$)

So:

$$ F = q\int_{S} \vec{E} \cdot \vec{n} \:\mathrm ds $$

$\endgroup$
1
$\begingroup$

No, because the surface charge density is not necessarily uniform.

In 2D, the amount of charge on a small line element $ds$ on the boundary is $\lambda ds$, where $\lambda$ is charge per length on the boundary. The force on it is then $\frac{1}{2}\lambda \vec{E} ds$. (The factor of $\frac{1}{2}$ arises from the fact that the net electric field "felt" by the charge on the surface is the average of the field inside the surface and the field outside the surface; and the field inside the surface is zero, since it's an equipotential inside.) If you integrate this over the whole boundary, you get the result $$ \vec{F} = \frac{1}{2} \int_S \lambda \vec{E} \, ds. $$ It's also worth noting that in 2D, $\vec{E} = (\lambda/\epsilon_0) \hat{n}$ near the surface of a conductor, and so you can figure out the charge density on the boundary if you know the electric field on the boundary (or vice versa.)

[Also, note that your proposed formula is dimensionally inconsistent. The right-hand side has units of (charge) * (electric field) * (area) — the last factor due to the integral — and so it doesn't have dimensions of force.]

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Sorry, if this is trivial, but would you mind explaining why I have to integrate $\sigma \vec{E}$ over the surface ? (and you would integrate over the surface, not the along the border line?) $\endgroup$ – james Apr 6 at 14:55
  • $\begingroup$ Sorry, I missed that this was a 2D problem. I'll revise things appropriately and add more detail. $\endgroup$ – Michael Seifert Apr 6 at 15:08
  • $\begingroup$ Thanks a lot ! A question: What is $\lambda$ in your formula ? $\endgroup$ – james Apr 6 at 15:20
  • $\begingroup$ @james: edited to clarify. $\endgroup$ – Michael Seifert Apr 6 at 16:45
  • $\begingroup$ Hmm... how would I know the charge per length ? I only have the electric field available... $\endgroup$ – james Apr 6 at 17:24
0
$\begingroup$

If the plates are close together, the force should be approximately (1/2)qE, because only half of the field between the plates comes from the other plate.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.