# Gauss law interpretation

#### Mondo

##### Junior Member
I stumbled upon this:
$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$This is a formula for an electric field of a sphere and the author makes a comment (Example 2.3 fundamentals of electrodynamics by Griffiths) "Notice a remarkable feature of this result: The field outside the sphere is exactly the same as it would have been if all the charge had been concentrated at the center." But how is that possible if in the formula above we clearly see that the electric field weakens as we move further from the sphere center proportionally to [imath]\frac{1}{r^2}[/imath](even intuitively it makes sense)

What is [imath]\hat r[/imath] in your formula? Do you, by chance, mean $\frac{1}{4\pi\epsilon_0}\frac{q}{r^3} \hat\mathbf r$, with [imath]r=|\hat\mathbf r|[/imath] ?

But how is that possible if in the formula above we clearly see that the electric field weakens as we move further from the sphere center proportionally to 1r2\frac{1}{r^2}r21(even intuitively it makes sense)
How is this different from the charge concentrated in the center?

I stumbled upon this:
$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$This is a formula for an electric field of a sphere and the author makes a comment (Example 2.3 fundamentals of electrodynamics by Griffiths) "Notice a remarkable feature of this result: The field outside the sphere is exactly the same as it would have been if all the charge had been concentrated at the center." But how is that possible if in the formula above we clearly see that the electric field weakens as we move further from the sphere center proportionally to [imath]\frac{1}{r^2}[/imath](even intuitively it makes sense)
When we want to calculate the electric field outside the sphere, we draw a Gaussian surface around the sphere. If the charge [imath]Q[/imath] is whether uniformly distributed inside the sphere or centered at the center, Gauss law only cares about the enclosed charge inside the Gaussian surface which is [imath]Q[/imath] in both cases. Therefore, you will get the same electric field outside the sphere for both scenarios.

On the other hand, things will be very different if we want to calculate the electric field inside the sphere.

Gauss law only cares about the enclosed charge inside the Gaussian surface
I know you are talking in the context of uniform distribution vs. charge in the center, but in the general case we can only make such statements about the field's flux through the surface.

I know you are talking in the context of uniform distribution vs. charge in the center, but in the general case we can only make such statements about the field's flux through the surface.
Professor blamocur, I don't really know if you agree or disagree with me! But you can try to solve the second scenario by yourself. Imagine that you have a sphere, it does not matter if it is insulating or conducting, of radius [imath]R[/imath] and has a charge [imath]Q[/imath] at its center. Calculate the electric field for [imath]r > R[/imath]. What do you get?

Professor blamocur, I don't really know if you agree or disagree with me! But you can try to solve the second scenario by yourself. Imagine that you have a sphere, it does not matter if it is insulating or conducting, of radius [imath]R[/imath] and has a charge [imath]Q[/imath] at its center. Calculate the electric field for [imath]r > R[/imath]. What do you get?
I was talking about the general case, i.e., when the charge is not necessarily in the center or when the distribution of the charge on a surface is not necessarily uniform. The flux equation still holds, but the specific formula might not. Sorry for the confusion.

What is [imath]\hat r[/imath] in your formula? Do you, by chance, mean $\frac{1}{4\pi\epsilon_0}\frac{q}{r^3} \hat\mathbf r$, with [imath]r=|\hat\mathbf r|[/imath] ?
@blamocur , as far as I know, the [imath]\hat{r}[/imath] represents a unit vector in the direction of [imath]r[/imath]. In the same way as in the carthesian coordinates you have [imath]f(x,y,z) = x\hat{x} + y\hat{y} + z\hat{z}[/imath]

When we want to calculate the electric field outside the sphere, we draw a Gaussian surface around the sphere. If the charge QQQ is whether uniformly distributed inside the sphere or centered at the center, Gauss law only cares about the enclosed charge inside the Gaussian surface which is QQQ in both cases. Therefore, you will get the same electric field outside the sphere for both scenarios.
Thanks for the response but I don't get it - the formula clearly shows that when we move further away from the center of the sphere the value of [imath]E[/imath] (electric field) decreases inversely proportional to a square of r.

@blamocur , as far as I know, the r^\hat{r}r^ represents a unit vector in the direction of rrr.
Looks correct in this case. I assumed that [imath]\hat r[/imath] stands for radius vector.

Thanks for the response but I don't get it - the formula clearly shows that when we move further away from the center of the sphere the value of [imath]E[/imath] (electric field) decreases inversely proportional to a square of r.
The electric field outside the sphere is like the electric field of a point of charge. It does not matter if the charge [imath]Q[/imath] will be at the center of the sphere or distributed uniformly when the sphere itself will be treated like a point of charge!

In other words, the sphere is just a big point of charge when you are calculating the electric field outside the sphere[imath] \ (r > R)[/imath].

The sphere and the point of charge have the same electric field formula:

Point of charge: [imath]\displaystyle E = k\frac{q}{r^2}[/imath]

Sphere: [imath]\displaystyle E = k\frac{Q}{r^2}[/imath]

It just the sphere usually has a bigger charge.

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Looks correct in this case. I assumed that [imath]\hat r[/imath] stands for radius vector.
So does the formula for electric field look correct to you? How would you explain my doubt then?

The electric field outside the sphere is like the electric field of a point of charge. It does not matter if the charge QQQ will be at the center of the sphere or distributed uniformly when the sphere itself will be treated like a point of charge!
@mario99 , but the formula clearly shows E depends on the distance. So are you saying the electric field of a point of charge is the same close to the charge as it is very far from it?

So does the formula for electric field look correct to you? How would you explain my doubt then?
[imath]\hat{r}[/imath] is just the direction. I am talking about the strength of the electric field (magnitude).

@mario99 , but the formula clearly shows E depends on the distance. So are you saying the electric field of a point of charge is the same close to the charge as it is very far from it?
not close but exactly the same!

I will give you an example. Let us assume that we have a point of charge, [imath]q = 10 \ \mu C[/imath], and we have a sphere of radius, [imath]R = 15 \ \text{cm}[/imath] that has the same charge at its center.

Calculate the electric field strength at a distance of [imath]20 \ \text{cm}[/imath] from the center of the sphere. Again calculate the electric field strength at a distance of [imath]20 \ \text{cm}[/imath] from the point of charge. Did you get the same electric field?

So does the formula for electric field look correct to you? How would you explain my doubt then?
Yes, it does look correct when [imath]\hat r[/imath] stands for the unit vector normal in the direction of the radius. As for your doubt, I am not sure what it is, but I am happy to try answering more specific questions.

What is [imath]\hat r[/imath] in your formula? Do you, by chance, mean $\frac{1}{4\pi\epsilon_0}\frac{q}{r^3} \hat\mathbf r$, with [imath]r=|\hat\mathbf r|[/imath] ?
Just for clarification, professor blamocur.

A unit vector [imath]\displaystyle \bold{\hat{r}} = \frac{\bold{r}}{r}[/imath].

So,

[imath]\displaystyle \Vert \bold{\hat{r}}\Vert = \frac{\Vert\bold{r}\Vert}{r} = \frac{r}{r} = 1[/imath].

[imath]\hat{r}[/imath] is just the direction. I am talking about the strength of the electric field (magnitude).

not close but exactly the same!

I will give you an example. Let us assume that we have a point of charge, [imath]q = 10 \ \mu C[/imath], and we have a sphere of radius, [imath]R = 15 \ \text{cm}[/imath] that has the same charge at its center.

Calculate the electric field strength at a distance of [imath]20 \ \text{cm}[/imath] from the center of the sphere. Again calculate the electric field strength at a distance of [imath]20 \ \text{cm}[/imath] from the point of charge. Did you get the same electric field?

Yes in this case it needs to be the same. The way I understood the comment in the book was the electric field on the sphere is say, X and then 3 miles away it is also X.. that would be, as author says "remarkable". But I guess when he says "it is the same as all the charge had been concentrated at the center" - he doesn't mean center of the sphere but rather a center as a point if sphere has not been there. In other words the electric filed 3 miles away from the shell of the sphere is the same as if we had calculated it having all the charge q at the center of coordinate system and a point 3 miles away from this center. Am I right?

he doesn't mean center of the sphere but rather a center as a point if sphere has not been there.
What's the difference? Where is your sphere located? You might consider making a drawing.
In other words the electric filed 3 miles away from the shell of the sphere is the same as if we had calculated it having all the charge q at the center of coordinate system and a point 3 miles away from this center. Am I right?
Does not sound right. The field 3 miles from a charged sphere would be the same as 6 miles from the same charge in the center of the sphere.

First, the OP has talked about the electric field outside the sphere, not on the sphere. Second, if the electric field on the sphere is [imath]X[/imath], and assuming [imath]R < 3 \ \text{miles}[/imath], the electric field at [imath]3 \ \text{miles}[/imath] will be weaker. Therefore, the electric field cannot be [imath]X[/imath] on both places. From the context of "concentrated at the center", it only means the center of the sphere.

In other words the electric filed 3 miles away from the shell of the sphere is the same as if we had calculated it having all the charge q at the center of coordinate system and a point 3 miles away from this center. Am I right?
From this sentence, I think that I understood what you are trying to say. You want to start the distance from the surface of the sphere. Then, you want to compare this with a point of charge.

Well when we have any spherical shapes, we will never be interested calculating the electric field starting at the surface of the sphere. (Electric fields are not calculated in that way.)

[imath]\displaystyle E = k\frac{Q}{r^2}, \ \ r[/imath] is the distance from the center.

Therefore, your reasoning does not make sense!

@mario99 , ok so it is about the same filed
Does not sound right. The field 3 miles from a charged sphere would be the same as 6 miles from the same charge in the center of the sphere.
That is exactly what does not make sense to me! How in the world the field can be the same? What about [imath]\frac{1}{r^2}[/imath] in the E formula? To me r is the distance from the charge/s creating the field - hence the electric field must weaken with distance there is no other way. How can you say otherwise?

@mario99 , ok so it is about the same filed
Electric fields created by spherical objects are calculated from their center. Even if you have a conducting sphere where all of its charges are located at its surface, the electric field it creates is calculated from its center.

@mario99 , ok so it is about the same filed

That is exactly what does not make sense to me! How in the world the field can be the same? What about [imath]\frac{1}{r^2}[/imath] in the E formula? To me r is the distance from the charge/s creating the field - hence the electric field must weaken with distance there is no other way. How can you say otherwise?
In my post 3 miles from (surface of) the sphere (of radius 3 miles) is 6 miles from the center of the center of that sphere. I.e., [imath]r[/imath] is the same 6 miles.