Electric field of a sphere

1. Sep 6, 2009

fluidistic

1. The problem statement, all variables and given/known data

The electric field on the Earth' surface is not null. Assume that the intensity of the field on Earth' surface is 300 N/C pointing in Earth's center. Calculate the excess of charges in Earth' surface giving the result in terms of electrons per $$m^2$$.

2. Relevant equations

Don't know.

3. The attempt at a solution

I'm stuck at starting the problem. How can the electric field point in Earth's center? Charges must be at the surface so I've no clue about the direction of the field.
I also do not have any idea about which formula to use.

Maybe $$E=A \cdot \sigma$$? So $$4 \pi r^2 \sigma= \frac{300N}{C} \Rightarrow \sigma =\frac{100}{6300^2 \cdot \pi}$$. But a dimensional analysis shows that I'm wrong. I get that the units of this are $$\frac{kg}{s^2Cm}$$ which clearly isn't a surface charge density. (I should get $$\frac{C}{m^2}$$ I guess.)

2. Sep 6, 2009

tiny-tim

Hi fluidistic!
Because the situation is spherically symmetric, so wherever you are, the field will be radial.

Hint: now draw a sphere around the earth.

3. Sep 6, 2009

fluidistic

Ah I see. Despite the fact that there's no charge inside the sphere (Earth), there can be an electric field. I still do not realize why the field lines are radials. I could imagine them to be different and still symmetric.

Hmm I don't see why.
I don't think I've to calculate the electric flux passing by Earth' surface.
Can you give me a little more hint?

4. Sep 6, 2009

tiny-tim

How?
You don't have to … but what harm could it do?

You know how much flux is going through the whole of the (outer) sphere, whatever its radius is, don't you?

5. Sep 6, 2009

fluidistic

See the picture I uploaded. I wrote a point "A", but do the same with all points on the sphere' surface. (Although I agree I did it for a circle but I'd keep the same idea for a sphere).

$$\Phi = \int _S Eda$$ where $$S$$ is the sphere' surface and $$E$$ is the electric field that I don't know.
$$da$$ would be an area element (I remember it's worth $$\rho ^2 \sin \phi d \rho d\phi d \theta$$)
But I think it can be simplified here, the flux would be $$4 \pi r^2 \cdot E$$ or something like that. But I don't know $$E$$.
I've also seen the formula $$\Phi = \frac{Q_ {\text{enclosed}}}{\varepsilon _0}$$ but $$Q _{\text{enclosed}}$$ is $$0$$... hence there's no flux passing through the sphere. It makes no sense! So I guess that $$Q$$ enclosed is not $$0$$.
I'm all confused as you can see.

File size:
2.3 KB
Views:
74
6. Sep 6, 2009

tiny-tim

Qenclosed (btw, isn't that easier than using LaTeX? ) by the outer sphere isn't zero …
… there are electrons all over the Earth's surface.

7. Sep 6, 2009

fluidistic

Ahh... I see. That's why you asked me to draw a sphere around the Earth.
I don't really know how to calculate $$Q _{\text{enclosed}}$$. I'm sure it's related to the intensity of the electric field on Earth' surface.
I give a try : Q is the sum of all the charges. Each charge is worth $$-1.6 \cdot 10^{-19}C$$.
I'm sorry I don't reach anything.

8. Sep 6, 2009

ideasrule

Do you know the formula for the electric field at a distance r from a point charge? A sphere is identical to a point charge in every respect, as long as we're not talking about the interior of the sphere. (That's why the formula F=GMm/r2 applies to planets, even though it assumes point masses.)

9. Sep 6, 2009

fluidistic

Oh thanks. Yes I know it ( I believe!), is it $$k \frac{q}{r^2}$$?

So for the sphere I replace $$q$$ by $$Q$$. But on the surface of the sphere, $$r=0$$ since you consider it as a point mass, right?
If so I'm lost.

10. Sep 6, 2009

ideasrule

No, r is the distance from the sphere's center, just as it is in the formula F=GMm/r^2. At the surface, r would be Earth's radius.

11. Sep 6, 2009

fluidistic

Ah ok! So I guess this mean that the electric field at the surface of the Earth is $$k\frac{Q}{(6300000 m)^2}$$. I can equal this to $$\frac{300N}{C}$$ (checking out the units of course) and I can get Q. Then I would divide it by $$4 \pi \cdot 6300000^2$$ in order to get the charge per meter squared. I would divide this result by the charge of the electron in order to reach the result.
I hope I'm not mistaken.

12. Sep 7, 2009

fluidistic

Thanks to both.
Out of curiosity, does the image I posted in the post #5 makes sense?
I mean, for me it's not natural to think that the electric field lines are radial in a sphere. I should let my intuition absorb this.

13. Sep 8, 2009

rl.bhat

Consider a small area Δs on the earth's surface. Consider a Gaussian surface in form of a cylinder enclosing this surface perpendicular to the earth's surface.
If ρ is the surface charge density, charge Δq on Δs = ρ*Δs.
One surface of the cylinder is inside the earth and the other surface is out side. So total flux is E*Δs = Δq /εο = ρ*Δs/εο
Hence E = ρ/εο . Now find ρ. The excess electron is given by ρ*A/e, where A is the area of the sphere and e is the charge on an electron.

14. Sep 8, 2009

tiny-tim

sorry, fluidistic, but i don't understand that image …

is that the surface of the sphere, or the interior of it?

either way, i don't see what's symmetrical about it

15. Sep 8, 2009

fluidistic

Ok thanks a lot.

It was the interior of a circle. The sketch is not complete, but would you forget about it ?:shy: