# Surface charge density

1. Sep 24, 2008

### ayestaran1

At a point where there's surface charge density sigma, by what amount is the electric field for a spherical shell of charge discontinuous?

E = 2*pi*k*sigma (although isn't that just true for a charged disc?)

2. Sep 24, 2008

### Defennder

Your problem is oddly stated. What does "by what amount" mean? Amount of what? Electric field? And what does it mean to say that the electric field is discontinuous for say, a given magnitude and direction?

3. Sep 24, 2008

### granpa

what is the electric field at that point without the extra charge density?

and do they mean point as in infinitesimal point?

4. Sep 24, 2008

### ayestaran1

Let me rephrase that:

The electric field for a spherical shell of charge is discontinuous by the amount [blank] at a point where there's a surface charge density sigma.

epsilon naught / sigma
sigma / epsilon naught
epsilon naught / sigma^2
epsilon naught 2 / sigma^2 (I think the book meant to say epsilon naught ^2 / sigma^2)
sigma^2 / epsilon naught

I'm really not sure what they're asking or how to figure this out.

5. Sep 25, 2008

### granpa

maybe they mean the electrc field ends at the surface. it goes from full strength to zero instantly and is therefore discontinuous.

6. Sep 25, 2008

### gabbagabbahey

The electric field is always discontinuous by an amount $$\sigma / \epsilon _0$$ when crossing a surface charge density of $$\sigma$$

The easiest way to show this is to use the fact that the electric field due to an infinite plane is $$\frac{\sigma}{2{\epsilon}_0} \hat{n}$$
where $$\hat{n}$$ is the unit vector perpendicular to the surface, pointing away from the surface. (this can be obtained using Gauss's Law) . If you are close enough to any surface charge, it will appear to be an infinite plane. The electric field will point in one direction on one side of the surface, and in the opposite direction on the other side of the surface and so the difference is:
$$\vec{E}_{above}-\vec{E}_{below}=\frac{\sigma}{2{\epsilon}_0} \hat{n}-\frac{\sigma}{2{\epsilon}_0} (-\hat{n})=\frac{\sigma}{\epsilon_0}\hat{n}$$

7. Sep 25, 2008

### ayestaran1

Thanks, but I'm still not sure what they're asking when they say "discontinuous by the amount..." - what does that mean? And how does that apply to a spherical shell?

My book defines the discontinuity of the electric field of an infinite disc/plane (I'm not quite sure what this means conceptually) as E = 2*pi*k*sigma = (2*pi) / (4*pi*epsilon naught) = sigma / (2*epsilon naught).

(so not simply sigma / epsilon naught)

What does this mean conceptually and how would it apply to a spherical shell?

8. Sep 25, 2008

### Defennder

Yeah I guess that's probably what they're referring to. The one described by gabba (his full name is too awkward and difficult to type out) works for a separate case of an infinite sheet.

9. Sep 25, 2008

### gabbagabbahey

The discontinuity should always be by an amount $$\sigma / {\epsilon}_0$$ . If your text gives a value of half that, then they must be referring to the difference between the value on one side of the surface, very very close to the surface and zero which I guess mean that the author of your text thinks the electric field right at the location of a charge is zero; which is incorrect (it is infinite).

For the difference between the field above and below a surface; you can check that you should get sigma/e_0: In the case of a constant charge density, the field inside a spherical shell of radius R, will be zero (you can check this using gauss's law) and the field immediately outside the surface will be
$$\frac{\sigma R^2}{\epsilon _0 r^2} |_{r=R}=\frac{\sigma}{\epsilon_0}$$.

10. Sep 25, 2008

### granpa

for a 2 dimensional plane of thickness dx the field above ond below is constant. it is independant of distance from the plane. why would the field be infinite AT the plane itself?

11. Sep 25, 2008

### Defennder

Did he say that the field at the plane itself is infinite? I think he was referring to a point charge.

12. Sep 25, 2008

### granpa

since the field of a surface isnt increasing as it approaches the surface I see no reason it would be infinite at the surface. even in the case of a point change why would it be infinite? if it is infinite then what direction does it point?

13. Sep 26, 2008

### gabbagabbahey

Well, for a point charge,
$$E \propto \frac{1}{r^2}$$
where r is the distance from the charge. What happens as r approaches zero?

The argument for the plane is a little different: if we let $$\hat{n}$$ be the unit normal upward froma horizontal plane charge, then the field(s) will be:
$$\vec{E}_{above}=\frac{\sigma}{{\epsilon}_0} \hat{n}, \quad \vec{E}_{below}=\frac{-\sigma}{{\epsilon}_0} \hat{n}$$
But what is the field at the charged plane is it $$\vec{E}_{above}$$ or $$\vec{E}_{below}$$??? The answer is that it is undefined aka $$\infty$$ . The average field on the other hand will be zero.

The fact that, classically, the electric field is infinite at the location of a charge is due the way in which the electric field is defined: "The electric field $$\vec{E}(\vec{r})$$ due to a collection of source charges is the force per unit charge experienced by a test charge located at $$\vec{r}$$".
How exactly is one to put a test charge at exactly the same location as the source charge?