# 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 ^^