When I need to calculate the electric field due to a charged infinite plate, does the intensity of the field not drop with distance from the plate?

Question

As opposed to calculating electric field due to a cylindrical wire or sphere where there is a dependence of the field to the distance between the point chosen and the charged body in itself. But in case of the charged plate, why do we not chose a reference point situated at a particular distance away from the plate where we need to find the electric field? Does this have something to do with the fact that the plate is infinite and no matter how far away we place the point from the plate, it is never too far away for the field to decrease?!

Answer 1

One way to realise that the field won't decrease it 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.

Answer 2

With a uniform charge density on an infinite plate (obviously not a realistic situation), you can choose a Gaussian surface at any distance from the plate since the field is uniform.

Answer 3

drawing a gaussian pillbox about a infinite surface charge

The field is symmetrical about any pillbox I chose as its infinite so the flux through it will be EA+EA the sides of the box contribute nothing as the field points perpendicular to the plate everywhere due to symmetry. thus 2EA =charge enclosed

charge enclosed is Surface charge density * area of pillbox

so 2EA=sigmaA 1/epsilon

E= sigma/2epsilon

which is independent of distance, so it is uniform no matter what for any location above or below my surface.

the reason this works is that for any box I pick, it will always be symetric due to the infinity.

This can also be confirmed with an actual integration using the electric field formula for a distribution

Where A is the area of a surface of my pilbox

edit: you are saying why is my gaussian surface not chosen as some reference point I want to evaluate my field at. IT IS! ITS its just that when you do this the height of my gaussiaj pillbox vanishes. meaning it's independent of that variable.

Answer 4

As with all Gaussian surface problems, this problem is about symmetry. Suppose the charged plane is the $x$-$y$ plane, and we want to find:

$${\bf E}(x,y,z)= E_x(x,y,z){\bf \hat x}+E_y(x,y,z){\bf \hat y}+E_z(x,y,z){\bf \hat z}$$

If we translate the coordinates in the plane:

$$ (x,y,z)\rightarrow (x',y',z')=(x+c_x,y+c_y, z) $$

the set-up of the problem doesn't change. It looks exactly the same.

Thus:

$${\bf E}(x',y',z)={\bf E}(x,y,z)$$

for all $c_x$ and $c_y$. The result cannot depend on translations parallel to the plane. Thus:

$${\bf E}(z)= E_x(z){\bf \hat x}+E_y(z){\bf \hat y}+E_z(z){\bf \hat z}$$

is sufficient.

What if you rotate about the $z$-axis?

$${\bf \hat x}'=\cos{\theta}{\bf \hat x}+\sin{\theta}{\bf \hat y} $$ $${\bf \hat y}'=-\sin{\theta}{\bf \hat x}+\cos{\theta}{\bf \hat y} $$

Since the physical set-up is unchanged by this rotation:

$$ E_{x'}(z)=E_{x}(z)$$ $$ E_{y'}(z)=E_{y}(z)$$

which is only solved for all $\theta$ by $E_x=E_y=0$. The electric field cannot pick out a preferred transverse direction.

Now we have:

$${\bf E}(z)=E_z(z){\bf \hat z}$$

The field must be parallel to the plane's normal, and can only depend on the distance to the plane.

One can reach that result in a coordinate-free manner: we need a vector $\bf E$, and we only have two vectors from which to construct it: the surface normal $\bf \hat n$, and the position $\bf r$. Thus we write the general equation:

$${\bf E}({\bf r}) = c_n{\bf \hat n}+ c_r{\bf r}+ c_p({\bf \hat n}\times{\bf r})+ c_z({\bf \hat n}\cdot{\bf r}){\bf \hat n}$$

where the $c_i$ are numbers.

Transverse and axial symmetries mean $c_r=0$. Parity symmetry means $c_p=0$; that is, under coordinate inversion it does not flip sign, while the vector electric field does flip sign.

That leaves $c_z$, which sits in front of the projection of position vector onto the surface normal, which is just the $z$-coordinate.

A non-zero $z$ dependence means there is a length scale to the problem. To get $c_z=0$, you need dilation symmetry.

Intuitively, you can understand there is dilation symmetry: no matter how you scale your distance from the infinite plane, it always looks the same. In this figure:

there are no tic marks on the axis, because there is no inherent scale. (One could also imagine a blackbody plane emitting light: no matter what distance you are from the plane, you see the same brightness...unless there is absorption, which introduces a length scale).

That argument is somewhat unsatisfying, because you could draw an infinite line of charge and make the same argument (though it would be erroneous).

Now we need to bring in units. We have two physical quantities at the start of the problem:

Surface charge density:

$$ \sigma_0 \sim \frac{\rm C}{\rm m^2} $$

and the permittivity of free space:

$$ \epsilon_0 \sim \frac{\rm m^3 kg}{\rm s^4 A^2} $$

where the units are standard S.I.units (one could also use dimensional math with $[D]$).

They combine in the problem as:

$$ \frac{\sigma_0}{\epsilon_0}\sim \frac{\rm m\cdot kg}{\rm s^3 A}$$

on the R.H.S., along with any powers of ${\rm m^{\alpha}}$ based on the $z$-coordinate, to produce and electric field:

$$ E \sim \frac{\rm m \cdot kg}{\rm s^3 A}$$

Clearly $\alpha=0$, so that $c_z=0$. There is dilation symmetry. Has we had a linear charge density ($\sim {\rm C/m}$), we would need $\alpha =-1$, introducing a $1/z$ dependence.

Finally:

$$ {\bf E}({\bf r}) = c_n{\bf \hat n}$$

is sufficient to describe the field.

You can find $c_n$ with symmetry: the field is equal an opposite on either side of the sheet; in one dimension the divergence is the derivative so:

$$ c_n - (-c_n) = \frac{\sigma}{\epsilon_0}$$