2
$\begingroup$

I understand why the field is perpendicular to the plate at all distances $R$ from the plate but can't understand why the strength remains the same. If we calculate $E$ using Gauss Law I understand we get a result that is independent of $R$ but how this result be different for a non-infinite plane? Basically what about the infinite plane causes the field to be uniform that that fails to occur in the finite version.

$\endgroup$

2 Answers 2

4
$\begingroup$

Here is a simplified intuitive picture:

No matter how far you are from the infinite plane, it looks the same. Zoom out, still infinite.

For a finite plane, the further you get, the smaller it looks. So the field gets weaker.

You can also do an integral. The electric field at distance $d$ perpendicular to the center of a circular plane will be proportional to:

$\int \frac{d}{(d^{2}+r^{2})^{3/2}} 2\pi r dr$

  • The $2\pi r dr$ factor represents a ring of charge at distance $\sqrt{d^{2}+r^{2}}$.
  • Another factor of $\frac{d}{\sqrt{d^{2}+r^{2}}}$ takes the perpendicular component of the electric field.
  • The remaining factor of $\frac{1}{d^{2}+r^2}$ is the inverse square distance.

Note that the derivative of $(d^{2}+r^{2})^{-1/2}$ is $-\frac{1}{2}(d^2+r^2)^{-3/2}2r$ so the integral becomes:

$-2\pi{}d(d^{2}+r^{2})^{-1/2} |_{r=0}^{r=R} = 2\pi - \frac{2\pi d}{\sqrt{d^{2}+R{^2}}}$

For an infinite plane ($R=\infty$) the integral becomes a constant ($2\pi$) that does not depend on the perpendicular distance $d$. For a finite plane, $d$ matters.

$\endgroup$
3
$\begingroup$

One way to realise that the field won't decrease for an infinite plate is to imagine the field lines.

The spacing between the field lines shows the strength of the field.

For a positive sphere all the lines point away from the centre. The field lines will get further apart as we go away from the sphere, so the field strength decreases.

For an infinite positive plate, by symmetry the lines must point away (perpendicular) to the plate. If we go further away, they'll stay equally spaced - this means that the field strength is independent of distance from the plate.

The difference between the infinite plate and the finite plate is that the symmetry is broken for the finite plate. The lines must not necessarily be perpendicular to the finite plate as there is now a reason to distinguish a preferred direction - i.e. all lines could point slightly away from the middle of the plate.

$\endgroup$
1
  • $\begingroup$ Rather than posting a close to identical answer to another one that that you have already posted, it is better to flag a question for closure as a duplicate. $\endgroup$
    – Buzz
    Nov 3, 2021 at 0:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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