Gauss' law

1. Nov 16, 2009

Cryphonus

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

A sphere of radius R=10 cm made of insulating material has a uniform charge density p=100 C/m-3

a)Find the magnitude of the electrical field at a distance r<R, using Gauss' Law.What is the value of the electric field at r=5 cm?
b)What is the total charge on this sphere?
c)Find the magnitude of the electric field at a distance r>R using Gauss' Law.What is the value of the electrical field at r=10cm? At r=20 cm?
e)Take the voltage at r=0 to be 0 Volts.Find the Voltage V(r) at any r by integrating the Electrical field E(r) along the radial direction.

2. Relevant equations

Gauss' Law= $$\oint$$E.ds=Qinside/E0 (permittivity constant)
3. The attempt at a solution

I managed to do b) since the density is provided to us we only have to find the Q from d= Q/V, V is 4/3Pi R3.... but i certainly do need help with other questions.

2. Nov 16, 2009

srmeier

I will start by helping you with part (a).

on the left hand side of Gauss' equation we are able to move the E field through the surface integral. why?

side note: (are you in calculus based physics or algebra based?)

3. Nov 16, 2009

Cryphonus

We are able to move the E field through the surface integral because the electric field is parallel with the surface vector of that area am i right?

(Calculus based physics)

4. Nov 16, 2009

srmeier

precisely!

now, we can modify our equation a little.

$$E\oint{dA}=\frac{Q_{en}}{\epsilon_0}$$

what is $$\oint{dA}$$ equal to?

5. Nov 16, 2009

Cryphonus

$$\oint{dA}$$ equals to surface area i assume?

6. Nov 16, 2009

srmeier

yes, now how would one express the charge in term of the charge density?

(remember that we desire an express in terms of the radius)

7. Nov 16, 2009

Cryphonus

well, charge density should be equal to total charge/Volume,,, and volume can be expressed as a function of the radius.

8. Nov 16, 2009

srmeier

right, so this is our expression:

$$E4\pi r^2=\frac{\rho 4 \pi r^3}{3\epsilon_0}$$

does this make sense to you?

Last edited: Nov 16, 2009
9. Nov 16, 2009

Cryphonus

To be honest i haven't seen this around before, would be glad if you can tell me the way that you calculated such a formula?

10. Nov 16, 2009

srmeier

We just calculated it together:

$$E$$= electric field
$$4 \pi r^2$$= surface area of a sphere =$$\oint dA$$
$$\rho=\frac{Q_{en}}{V_s}$$ (where V is the volume of the sphere.)

so $$\frac{\rho 4 \pi r^3}{3}=Q_{en}$$

what doesn't make sense?

11. Nov 16, 2009

Cryphonus

right! i got it now. its all clear now :)

12. Nov 16, 2009

Cryphonus

So how are we going to plug this to the question. It feels like there is still things that im missing ?

13. Nov 16, 2009

srmeier

(sorry I went out to eat, I am back now for 15mins then I have class just f.y.i)

nothing is missing just plug in what is known and solve for the E field.

Note that this is the E field for only 0<r<R

14. Nov 16, 2009

Cryphonus

thats ok =). I have a question about that.What changes when we take a point that is greater than our radius, and when we take a point that is smaller than the radius, which is actually within the sphere.

15. Nov 16, 2009

srmeier

The charge enclosed changes from linear to constant.

When: 0<r<R

the charge enclosed is a function of the radius. This is quite intuitive because the closer we get to R the more charge we have enclosed.

When: R<r<infinity

the charge enclosed is a constant function. This is also quite intuitive because there is no charge outside the sphere which would add to the enclosed charge.

16. Nov 16, 2009

Cryphonus

Ok that makes sense except 2 points;

First how are we going to find the formula for the other situation, what do we need to change?

Second one;

For the situation, R<r<infinity , this actually means the point we are considering is somewhere out of the sphere.Are you saying that the E field is constant for all the points that is left out of the sphere?, shouldn't it be decreasing with proportional to 1/d2

17. Nov 16, 2009

srmeier

we are going to find the formula for the other situations in the same manner. We need to change the charges dependence on r.

point two:
I am not saying the E field is constant outside of the sphere. I am saying that the enclosed charge is constant outside of the sphere. You should be able to differentiate between these two things.

(Note: the E field will fall off inversely proportional to the radius squared)

Last edited: Nov 16, 2009
18. Nov 17, 2009

Cryphonus

Alright i understood.Thanks for all the help

19. Nov 17, 2009

srmeier

My pleasure ^^