# Uniform Field due to an Infinite Sheet of Charge

• Zahid Iftikhar
In summary, Zahid Iftikhar is asking for help understanding how an electric field due to an infinite sheet of charge can be uniform where ever you experience it. He is not convinced by the argument that the field intensity depends on distance from the sheet of charge. He argues that the field intensity should decrease as the inverse square but the amount of charge increases as the square, so there is no net change in force from any infinitesimal solid angle. Additionally, he suggests that the system be treated with Gauss's Law and the symmetry argument due to the infinity and the symmetry.

#### Zahid Iftikhar

Hi
I have read one excellent explanation on the question I am posting again, at another thread on PF, because I am not able to completely understand the concept. Please spare some more time.
I need help to understand how an electric field due to an infinite sheet of charge can be uniform where ever you experience it. I know the electric fields die down inverse squarely by increasing the distance. I understand all the charges keep the field perpendicular and straight from the plane of the sheet. But it makes me confused to believe that if a charge is placed at 1cm distance and then moved to 4cm distance along the same electric field line, there will be same force. In this way if a charge is taken to infinity, again there will be same field intensity. The equation, I know , for electric field intensity is independent of distance from the sheet of charge but physically it seems incorrect.
As far this explanation goes that components of E parallel to the plane of sheet cancel out and those perpendicular add up to produce uniform field does not convince me. Here is the reason.
For a point near the sheet, the supporting components of E will have greater strength, so they will produce strong field near the sheet. Whereas for a point away from the sheet of charge, these components should have individually weak values and thus when add should produce weak field.
High regards
Zahid Iftikhar

• Delta2
Zahid Iftikhar said:
Whereas for a point away from the sheet of charge, these components should have individually weak values
You are just making an assumption here because it 'seems right' to you. What you say is ok for a 'small' charged plate and it will tend to inverse square law at a large distance but it is not right for an infinite plate.

• Delta2
Have you done the calculation using a force balance on the charge, or have you done it using Stokes' law?

Zahid Iftikhar said:
For a point near the sheet, the supporting components of E will have greater strength, so they will produce strong field near the sheet. Whereas for a point away from the sheet of charge, these components should have individually weak values and thus when add should produce weak field.
This is an incomplete quantitative argument. You simply need to work through the math to show that it is incorrect.

However, there is a very simple qualitative argument by symmetry. In order for the field to decrease, the field lines would need to diverge. If the field lines were to diverge then there would need to be some special point on the plane which is at the center of the divergence. However, since the plane is infinite there is no justification for picking any such point. Therefore, by symmetry there can be no divergence of field lines and thus no decrease in field strength.

• Delta2
Dale said:
This is an incomplete quantitative argument. You simply need to work through the math to show that it is incorrect.
The reason that the argument is incorrect (as illustrated by the mathematics) is that, as the test charge is moved higher, the directionality of the electrostatic forces from all points on the plane become more vertical. This exactly cancels out the effect of the greater distance.

• Delta2
Chestermiller said:
The reason that the argument is incorrect (as illustrated by the mathematics) is that, as the test charge is moved higher, the directionality of the electrostatic forces from all points on the plane become more vertical. This exactly cancels out the effect of the greater distance.
Yes, or another way to think of it is that, regardless of the distance, the plane subtends exactly 2 pi steradians of solid angle. For every infinitesimal solid angle, as the distance increases the force per unit charge decreases as the inverse square but the amount of charge increases as the square, so there is no net change in force from any infinitesimal solid angle. Combined with no change in solid angle, the force is constant.

• Nathanael
What you saying might have some good intuition in it however it is of the cases where our intuition is wrong. The reasons are explained by Chet and Dale in their posts. The only thing I have to add is that this system is better to treat with Gauss's Law and the symmetry argument (due to the infinity and the symmetry we can conclude that the field can have only one possible direction and that is vertical to the infinite sheet, from this conclusion and from Gauss's Law we can infer that the field magnitude will be the same everywhere , independent of the distance) , instead of the inverse square law.

Just to state something which you might take as a guideline, "Whenever there is some sort of infinity involved in a system, there is a good probability that we have some counter intuitive results regarding the behaviour of the system".

• Dale
Thanks all for your kind replies.
I have found one explanation that seems quite intuitive.
For a point close to the sheets, the components of electric intensity, (sine components) parallel to the plane of sheet will cancel. The cosine components will add together but their individual values will be small for the angle will be large. At a farther point, the angle of E due to neighboring charges gets smaller, consequently the cosine components have greater values and provide compensation of decrease in E that should have been there due to greater distance of the second point. #### Attachments

• sophiecentaur
Chestermiller said:
Have you done the calculation using a force balance on the charge, or have you done it using Stokes' law?
Thanks Sir. I am trying to understand it physically using little mathematics.

Dale said:
Yes, or another way to think of it is that, regardless of the distance, the plane subtends exactly 2 pi steradians of solid angle. For every infinitesimal solid angle, as the distance increases the force per unit charge decreases as the inverse square but the amount of charge increases as the square, so there is no net change in force from any infinitesimal solid angle. Combined with no change in solid angle, the force is constant.
Thank you Sir.

Delta² said:
Just to state something which you might take as a guideline, "Whenever there is some sort of infinity involved in a system, there is a good probability that we have some counter intuitive results regarding the behaviour of the system".
Very good lines indeed.
Regards

• Delta2
Zahid Iftikhar said:
Thanks Sir. I am trying to understand it physically using little mathematics.
That's what I was alluding to in post #5.

• Delta2
Just to note something, it is not only what @Chestermiller states in post #5, and OP states similar in post #8, it is that together with the infinity that make the electric field independent of distance.

To see it clearly let's write down some math for this situation, working in a cartesian coordinate system and with the inverse square law:

Lets take an infinitesimal area dS in the plane z=0, which is dS=dxdy and is at point (x,y). The charge of dS is simply ##dq=\rho dxdy## where ##\rho## the surface charge density, and according to the inverse square law the magnitude of infinitesimal electric field from this dq at a point ##(x_0,y_0,h)## which is at distance ##h## above the infinite charged plane is

##dE=\frac{\rho dxdy}{(x-x_0)^2+(y-y_0)^2+h^2}##

The direction of this force makes an angle ##\theta## with the horizontal plane such that ##\sin\theta=\frac{h}{\sqrt{(x-x_0)^2+(y-y_0)^2+h^2}}##. Here we see what post #5 and #8 note , that ##\sin\theta## increases as ##h## increases and is 1 only when ##h## is infinite.

So the vertical component of ##dE## is ##dE_z=dE\sin\theta=\frac{\rho hdxdy}{{((x-x_0)^2+(y-y_0)^2+h^2)}^{\frac{3}{2}}}##.

Thus to find the total vertical component of E we simply sum the infinitesimals ##dE_z##, that is we integrate in this case:

##E=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\frac{\rho h}{{((x-x_0)^2+(y-y_0)^2+h^2)}^{\frac{3}{2}}}dxdy## (1)

This integral according to Wolfram http://www.wolframalpha.com/input/?i=integral+h/((x-a)^2+(y-b)^2+h^2)^(3/2)dxdy+from+x=-infinity+to+x=infinity+and+from+y=-infinity+to+y=infinity is independent of ##x_0,y_0## and essentialy independnt of ##h## and it is equal to ##\frac{2\pi \rho h}{|h|}## where the ##\frac{h}{|h|}## gives us essentially the sign which says if the direction of the field will be up or down , depending if we get the point up or down from the surface (h positive or negative respectively).

if the intervals of integration for x,y in (1) are not from ##-\infty## to ##+\infty## then the integral will depend on ##x_0,y_0## and h as well.

I shall post more details later on how to calculate the integral using successive integration on x,y.

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First finding the antiderivative of E with respect to x we get ( I omit in the following the factor off ##\rho h## in the numerator which is a constant and can be taken out of the integral, I ll add it in the very end) ##F_x(x,y)=\frac{x-x_0}{((y-y_0)^2+h^2)\sqrt{(x-x_0)^2+(y-y_0)^2+h^2}}##

thus integrating first with respect to x , we get ##f(y)=\lim_{x\rightarrow+\infty}{F_x(x,y)}-\lim_{x\rightarrow-\infty}{F_x(x,y)}##.

Lets first look at ##\lim_{x\rightarrow+\infty}{F_x(x,y)}## we can see that after some algebra it is essentially equal to
##lim_{x\rightarrow+\infty}\frac{x-x_0}{((y-y_0)^2+h^2)|x-x_0|\sqrt{1+\frac{(y-y_0)^2}{(x-x_0)^2}+\frac{h^2}{(x-x_0)^2}}}=\frac{1}{((y-y_0)^2+h^2)\sqrt{1+0+0}}##

Similarly ##\lim_{x\rightarrow-\infty}{F_x(x,y)}=-\frac{1}{(y-y_0)^2+h^2}##.

So ##f(y)=\frac{2}{(y-y_0)^2+h^2}## so we see how taking the limits for x to infinity removes the dependence on ##x_0##

Proceeding with the integration for y, we got to find the antiderivative of ##f(y)## and then take the limits to +-infinity.

It is ##F(y)=\frac{2\arctan\frac{y-y_0}{h}}{h}## and for the final value of integral we got to take

##E=\lim_{y\rightarrow+\infty}{F(y)}-\lim_{y\rightarrow-\infty}{F(y)}##

We can easily see that those limits are the first ##2\frac{\pi}{2|h|}## and the second ##-2\frac{\pi}{2|h|}## so (putting back the factor ##\rho h## which I omit at star
)
##E=\rho h4\frac{\pi}{2|h|}=2\pi\rho\frac{h}{|h|}##. So we see how taking the limits of y to infinity removes the dependence on ##y_0##. All it remains is a technical dependence on h which is the sign of h ##sgn(h)=\frac{h}{|h|}##.

One can see that if we don't take limits to infinity but instead limits to some finite boundaries ##a,b## (for x and y) then we will not be able to get rid of ##x_0## and ##y_0## and in the denominator we would have something that involves ##x_0##, ##y_0## and ##h^2## as well as ##a## and ##b## , so the dependence on h wouldn't take the simple form of ##\frac{h}{|h|}##.

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Delta² said:
First finding the antiderivative of E with respect to x we get ( I omit in the following the factor off ##\rho h## in the numerator which is a constant and can be taken out of the integral, I ll add it in the very end) ##F_x(x,y)=\frac{x-x_0}{((y-y_0)^2+h^2)\sqrt{(x-x_0)^2+(y-y_0)^2+h^2}}##

thus integrating first with respect to x , we get ##f(y)=\lim_{x\rightarrow+\infty}{F_x(x,y)}-\lim_{x\rightarrow-\infty}{F_x(x,y)}##.

Lets first look at ##\lim_{x\rightarrow+\infty}{F_x(x,y)}## we can see that after some algebra it is essentially equal to
##lim_{x\rightarrow+\infty}\frac{x-x_0}{((y-y_0)^2+h^2)|x-x_0|\sqrt{1+\frac{(y-y_0)^2}{(x-x_0)^2}+\frac{h^2}{(x-x_0)^2}}}=\frac{1}{((y-y_0)^2+h^2)\sqrt{1+0+0}}##

Similarly ##\lim_{x\rightarrow-\infty}{F_x(x,y)}=-\frac{1}{(y-y_0)^2+h^2}##.

So ##f(y)=\frac{2}{(y-y_0)^2+h^2}## so we see how taking the limits for x to infinity removes the dependence on ##x_0##

Proceeding with the integration for y, we got to find the antiderivative of ##f(y)## and then take the limits to +-infinity.

It is ##F(y)=\frac{2\arctan\frac{y-y_0}{h}}{h}## and for the final value of integral we got to take

##E=\lim_{y\rightarrow+\infty}{F(y)}-\lim_{y\rightarrow-\infty}{F(y)}##

We can easily see that those limits are the first ##2\frac{\pi}{2|h|}## and the second ##-2\frac{\pi}{2|h|}## so (putting back the factor ##\rho h## which I omit at star
)
##E=\rho h4\frac{\pi}{2|h|}=2\pi\rho\frac{h}{|h|}##. So we see how taking the limits of y to infinity removes the dependence on ##y_0##. All it remains is a technical dependence on h which is the sign of h ##sgn(h)=\frac{h}{|h|}##.

One can see that if we don't take limits to infinity but instead limits to some finite boundaries ##a,b## (for x and y) then we will not be able to get rid of ##x_0## and ##y_0## and in the denominator we would have something that involves ##x_0##, ##y_0## and ##h^2## as well as ##a## and ##b## , so the dependence on h wouldn't take the simple form of ##\frac{h}{|h|}##.
This is much easier to do in cylindrical coordinates.

Chestermiller said:
This is much easier to do in cylindrical coordinates.

##\int_0^{\infty}\int_0^{2\pi}\frac{\rho rh}{(r^2+r_0^2+h^2+2rr_0\cos(\phi-\phi_0))^{3/2}}drd\phi##

Is that a lot easier to calculate?? Be my guest Chet :D

Delta² said:
##\int_0^{\infty}\int_0^{2\pi}\frac{\rho rh}{(r^2+r_0^2+h^2+2rr_0\cos(\phi-\phi_0))^{3/2}}drd\phi##

Is that a lot easier to calculate?? Be my guest Chet :D
If the test charge is at z = h, and, considering the ring of charge ##2\pi r \rho dr## between r and r + dr, the upward force on the test charge q is given by: $$dF=\frac{1}{4\pi\epsilon_0}\frac{q(2\pi r \rho dr)}{(r^2+h^2)}\frac{h}{\sqrt{r^2+h^2}}=\frac{1}{2\epsilon_0}\frac{ \rho hrdr}{(r^2+h^2)^{3/2}}q$$

Chestermiller said:
If the test charge is at z = h, and, considering the ring of charge ##2\pi r \rho dr## between r and r + dr, the upward force on the test charge q is given by: $$dF=\frac{1}{4\pi\epsilon_0}\frac{q(2\pi r \rho dr)}{(r^2+h^2)}\frac{h}{\sqrt{r^2+h^2}}=\frac{1}{2\epsilon_0}\frac{ \rho hrdr}{(r^2+h^2)^{3/2}}q$$

You consider the test charge at ##(0,0,h)## (if I understand correctly). What if it is at ##(r_0,\phi_0,h)##?

EDIT :Ok I guess you going to invoke some symmetry arguments, but my approach does it without invoking symmetry :D

Delta² said:
You consider the test charge at ##(0,0,h)## (if I understand correctly). What if it is at ##(r_0,\phi_0,h)##?
Why I would ever choose to do this? It just makes the analysis desperately more complicated with no advantage.

Incidentally, the relationship I gave readily integrates analytically to $$F=\frac{\rho q}{2\epsilon_0}$$
So the field strength is given by: $$E=\frac{\rho}{2\epsilon_0}$$
This is consistent with the result obtained more simply by applying Gauss' Law.

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• nasu