Why does the plate have to be infinite?

[Moazin Khatri]

When I was taught Gauss's law. My teacher used a cylindrical Gaussian surface to find the electric field above an infinite uniformly charged plate. What I have trouble understanding is why the plate has to be infinite in order for the arguments to work? http://farside.ph.utexas.edu/teaching/302l/lectures/node27.html Here's a link explaining the way to use Gauss's law to find the electric field. But what if I repeat the same thing with a finite plate? I know it won't work but I want a good explanation of why it won't work?

 

[Alpharup]

yess...it is possible for finite plate but it involves integration.

 

[Moazin Khatri]

Yes I know we can use calculus to find electric field both above an infinite or finite plate. But I want a good argument for why the method the link follows is only applicable to infinite plate? I can take a finite plate and still follow the same process.

 

[Alpharup]

I think we can give an intutive explation. By gauss law , charge enclosed inside closed surface= €×(sum of(Surface Area vector× Electric Field vector)). Let us consider a point charge q.( By point charge, we ideally assume it is so so small...). The field is a function of q. Draw a sphere of radius R. the area is (4*pi*(R^2)).

The electric field is (q/4*pi*€*R^2). Multiply both Field and area( Intutively, the field vector is along area vector). So, the dot product multiplication results in a scalar q/€. The scalar is electric flux.

A similar argument can be said of a large rock carrying lot of charge. You take your vehicle and travel a long distance, such that the rock appears tiny? Cant we use point charge approximation?
You told that you have trouble understanding infinite plate argument. Let the radius of plate be 50 m. Ant ant is near the centre of the plate. For the ant, only the plate is visible in the vicinity. Wheverever it goes, it always ends up in the plate. Why can't the ant consider it as infinite plate. Approximations are good as long as it does not affect what we require.

 

[Merlin3189]

we expect the electric field on either side of the plane to be a function of

only, to be directed normal to the plane

That is because the plate is infinite.
If the plate were finite, then the field would not be normal to the plane. It would diverge slightly. This is obvious near the edge, but must also be true (at least a tiny bit) everywhere else except at the exact centre of the plane (and only at an infinitesimal point there.)
This ideal assumption avoids the complication of having to account for the slight divergence of the field, so makes a nice neat formula.

later they consider a parallel plate capacitor and use the simple formula,

Outside this region, the electric field cancels to zero. The above result is only valid for two charged planes of infinite extent. However, the result is approximately valid for two charged planes of finite extent, provided that the spacing between the planes is small compared to their typical dimensions.

This sort of 'trick' is often used to avoid very complex calculations and producing cumbersome overcomplex formulae.

Edit: The comment about, "spacing is small compared to their typical dimensions" is essentially the "ant" argument of sharan swarup.

 

[Moazin Khatri]

Yes. I got it. Thanks a lot Sharan swarup and Merlin.