We know that electric field due to an INFINTE large sheet is constant and at INFINTY the electric field is not zero. But say if I take a finite sheet of length l and width w. Then the electric field would be zero at infinity. Also we say that electric field is constant for a infinite sheet but for a finite sheet of length l and width w , it won't be constant. Electric field for a finite long sheet at a point x above it whose length is l and width is w is given in th following website:
http://people.rit.edu/jdasps/jdainfo/313_tp/UniformlyChargedFinitePlane.pdf
Just give a look where E finite plane is written in tha website….in the second page ….And that formula is in box. When you put x as infinity you would make arctan tending towards zero and hence the you would make the electric field when x is tending to infinity as zero. So the formula E=sigma/2ϵ0 (which is constant) is only for the plane sheet which is infinitely large and for this infinitely large sheet of charge at infinty the field is same i.e. E=Sigma/2ϵ0. But for a finite sheet at infinty the electric field is zero and also for a finite sheet the electric field is not constant with changing the distance from the plate. Then in the case of capacitors, whose plates are finite, why then we say that electric field due to one of its plate is sigma/2ϵ0? So my first question: it right to say that for any plane sheet of charge which is not infintely large electric field is not constant and at inifinty the electric field is zero? And my second question: plz explain why we say that electric field due to one of the capacitor plate is sigma/2ϵ0 while the capacitor plates are finite.
[Physics] Why is this electric field due to one plate of a capacitor $\sigma / 2 \epsilon_0$ when the capacitor plates are finite
capacitanceelectric-fieldselectrostatics
Related Solutions
When discussing an ideal parallel-plate capacitor, $\sigma$ usually denotes the area charge density of the plate as a whole - that is, the total charge on the plate divided by the area of the plate. There is not one $\sigma$ for the inside surface and a separate $\sigma$ for the outside surface. Or rather, there is, but the $\sigma$ used in textbooks takes into account all the charge on both these surfaces, so it is the sum of the two charge densities.
$$\sigma = \frac{Q}{A} = \sigma_\text{inside} + \sigma_\text{outside}$$
With this definition, the equation we get from Gauss's law is
$$E_\text{inside} + E_\text{outside} = \frac{\sigma}{\epsilon_0}$$
where "inside" and "outside" designate the regions on opposite sides of the plate. For an isolated plate, $E_\text{inside} = E_\text{outside}$ and thus the electric field is everywhere $\frac{\sigma}{2\epsilon_0}$.
Now, if another, oppositely charge plate is brought nearby to form a parallel plate capacitor, the electric field in the outside region (A in the images below) will fall to essentially zero, and that means
$$E_\text{inside} = \frac{\sigma}{\epsilon_0}$$
There are two ways to explain this:
The simple explanation is that in the outside region, the electric fields from the two plates cancel out. This explanation, which is often presented in introductory textbooks, assumes that the internal structure of the plates can be ignored (i.e. infinitely thin plates) and exploits the principle of superposition.
The more realistic explanation is that essentially all of the charge on each plate migrates to the inside surface. This charge, of area density $\sigma$, is producing an electric field in only one direction, which will accordingly have strength $\frac{\sigma}{\epsilon_0}$. But when using this explanation, you do not also superpose the electric field produced by charge on the inside surface of the other plate. Those other charges are the terminators for the same electric field lines produced by the charges on this plate; they're not producing a separate contribution to the electric field of their own.
Either way, it's not true that $\lim_{d\to 0} E = \frac{2\sigma}{\epsilon_0}$.
Toby, I agree that this is really counter intuitive and I was also quite surprised as well when I first saw this very demonstration. I am an undergraduate TA and this is how I explained it in my lab section. I hope this helps. I see two parts to a full explanation: (1) Why is the electric field constant and (2) why does the potential difference (or voltage) increase?
Why is the electric field constant as the plates are separated? The reason why the electric field is a constant is the same reason why an infinite charged plate’s field is a constant. Imagine yourself as a point charge looking at the positively charge plate. Your field-of-view will enclose a fixed density of field lines. As you move away from the circular plate, your field-of-view increases in size and simultaneously there is also an increase in the number of field lines such that the density of field lines remains constant. That is, the electric field remains constant. However, as you continue moving away your field-of-view will larger than the finite size of the circular plates. That is, the density of field lines decreases and therefore, the electric field decreases as well as the potential field.
To show this mathematically, the easiest way to show this for E = constant is using the relation between the electric and potential fields: $$E = -\frac{\Delta V}{\Delta d} \longrightarrow \Delta V =-E \Delta d$$ I would expect the voltage to increase linearly as long as the field is constant. When the electric field starts decreasing, the voltage also decreases and the fields behave as finite charged plates. Although I’ve only talked about one plate, this idea immediately applies to two plates as well.
Why does the work increase the electrical potential energy of the plates? One way to interpret why the voltage increases is to view the electric potential (not the electrical potential energy) in a completely different manner. I think of the potential function as representing the “landscape” that the source (of the field) sets up. Let me explain what the gravitational potential acts like when a ball is thrown upwards (of course, you know what happens in terms of the force of gravity or in the conservation of energy scenario). I claim that the potential function is related to the “gravitational landscape” that the earth sets up, which is derived from the potential energy and is equal to the potential energy per mass:
$$ {\Delta U = mg\Delta y} \longrightarrow \frac{\Delta U}{m} = \Delta V = g\Delta y$$
Plotting these functions, the constant gravity field sets up a gravitational potential ramp (linear behavior) that looks like
In terms of energy, the ball moves up this gravitational ramp were the ball is converting its kinetic energy into potential energy until the ball reaches it maximum height. However, the gravitational ramp exists whether the ball is thrown up or not. That is, gravity sets up a gravitational ramp (the landscape) and this is what the ball “sees” before it is thrown up.
If we now apply the above thinking to a constant electric field between the parallel plates, the electric potential function is derived in a similar manner:
$$ {\Delta U = qE\Delta r} \longrightarrow \frac{\Delta U}{q} = \Delta V = E\Delta r$$
If we look at the electric potential of the negative plate (it’s easier than the positive plate), it has a negative electrical ramp that starts at 0V.
So as your TA pulls the plates apart, the work she does moves the positive plate up the electrical ramp and increases the potential of the positive plate. So this interpretation of the electric potential is what you intuitively already think about in terms of mechanical situations like riding your bike up a hill. There is no difference in the electrical situation.
Best Answer
Yes it is right to say that electric field by a finite plate is not constant and zero at infinity. But in case of capacitors,the separation between the plates is so small as compared to dimensions of plate that with respect to the separation between the plates the plate itself can be considered as infinite. It is just a relative assumption to simplify things. Take care :-)