How does the electric field strength due to an arrangement of infinitely charged plates remain constant?

Question

For a point charge, the electric field is given something as the inverse of the square of the distance from it. But, that only means that as we go farther away from the point charge, the number of electric field lines PER UNIT AREA, will decrease. Therefore, if a test charge were to be placed at a far away distance, it would experience a lesser force (as there are lesser number of electric field lines per unit area) as opposed to one placed near the point charge.

The reason why I asked this question was because I was confused when I read the statement that 'electric field strength produced due to an arrangement of two infinite sheets/plates is independent of the distance from the plates.' I mean, if we consider the electric field due to ONE of the charges on the plate, then it must be dependent on distance, but it isn't.. so the only suitable reason I could think of is that along an electric field line, no matter where ON IT, the electric field strength due to that ONE line, will remain constant.

Am I wrong?

Answer 1

so the only suitable reason I could think of is that along an electric field line, no matter where ON IT, the electric field strength due to that ONE line, will remain constant.

You can't think about the electric field strength as being "due to a line". You have to think of the electric field strength in terms of the density of the lines.

The density of the electric field lines between the two parallel plates is constant, and thus the field strength is constant, because the field of a single charge on a plate is added to fields produced by all the other surrounding charges on the plate to make the overall field, and density of field lines, uniform between the plates.

See the answers to the question posed in the following link:

Proving electric field constant between two charged infinite parallel plates

Hope this helps.

Answer 2

The mathematical answer to your question is to derive Gauss's law from Coulomb's law (https://en.wikipedia.org/wiki/Gauss%27s_law#Deriving_Gauss's_law_from_Coulomb's_law), then apply Gauss's law to the infinite sheet (http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html).

For an intuitive answer, consider an infinite sheet of charge with a pinhole in it. Suppose we start a charge in that pinhole. All the electric forces pushing on the charge in the pinhole are counterbalanced by an equal and opposite push from the opposite side, so the charge experiences a net force of 0. We know the $E \propto F$ so E must also be 0.

Now suppose we move the charge just the tiniest bit upwards. The net force in any direction other than up is still 0, but now each charge contributes an upwards force $F \propto \frac{sin(\theta)}{r^2}$.

$sin(\theta) \approx 0$ for $\theta \approx 0$ and $sin(\theta) \approx 1$ for $\theta \approx \pi/2$

Let's pick some angle $0 \lt x \lt \pi/2$ and approximate all the distant charges $\theta \lt x$ as $\theta = 0$ and pretend all the nearby charges $\theta \gt x$ are on a point directly under our test charge.

In theory we could pick x just right (by doing calculus and finding the average sin component provided by the infinite distribution), but let's just suppose we drastically over-estimate for now and say $x = 10^{-1000000}$ or so.

Now we've killed our infinity by multiplying it by 0 and what we're left with is a point charge a distance r away from a "source" point charge $Q$ with the magnitude of all the charges under the cone defined by angle x and height r.

Coulomb's law tells us how to solve this. For a point charge $Q$, $$E \propto \frac{Q}{r^2}$$

But as we move away, our cone of height r and angle $\theta = x$ sweeps out more and more charges on the surface, starting at 0 when r = 0. The base of a cone varies according to the square of the height of the cone, therefore $Q \propto r^2$

So, $$E \propto \frac{Q}{r^2} \propto \frac{r^2}{r^2} \propto 1$$

So, E is invariant with r.