# Homework Help: EM field between 2 superimpose spheres

1. Aug 24, 2009

### Luchopas

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

Problem:

2 spheres each one of them with a radius R and uniformly charged with ro+ and ro-, are situated in a way they superimpose partially ( see figure). Let be "d" the vector from the positive center to the negative center.

http://img220.imageshack.us/img220/9633/diagrame.png" [Broken]

2. Relevant equations

Find the electric field in the hollow section.

3. The attempt at a solution

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

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 4, 2017
2. Aug 24, 2009

### kuruman

Hi Luchopas, welcome to PF. Pretend that there is only one sphere. Can you find the E field in the region r < R for the other sphere?

3. Aug 24, 2009

### gabbagabbahey

Hi Luchopas, welcome to PF!

Hint: what is the electric field $\textbf{E}(\textbf{r}_{\pm})$ inside a uniformly charged sphere of radius $R$ and charge density $\pm\rho$, where $\textbf{r}_{\pm}$ is the vector from the center of the positively/negatively charged sphere to the field point?

P.S. This forum supports $\LaTeX$. For an introduction to using it on these forums, click here. You can also click on ay of the above $\LaTeX$ images to see the code that generated them.

4. Aug 24, 2009

### Luchopas

sorry , dind't catch your question , i can find the electric filed of one sphere and the other... but not the electric field , in the hollow section...

for a solid sphere the elctric field inside and outside is

E= (1/ 4*pi*epsilon0) * ro/r-squared ( in r direction)

for the other sphere is the same but with negative ro

but after this is where i fail...

5. Aug 24, 2009

### gabbagabbahey

No,

$$\textbf{E}=\frac{1}{4\pi\epsilon_0}\frac{\frac{4}{3}\pi R^3\rho}{r^2}\hat{r}$$

is the field outside (I'm assuming that typing ro instead of ro*volume was a typo on your part?), but not the field inside. What is the field inside?

Last edited: Aug 24, 2009
6. Aug 24, 2009

### Luchopas

mm how do you get t this calculation???
in the inside would be the same right?
do you have any msn messenger to talk faster?
thanks

p.D: sorry im new in this physics subject

7. Aug 24, 2009

### gabbagabbahey

No, the field inside is different.

The calculation is usually done as an example in most introductory EM texts...Which text are you using?

8. Aug 24, 2009

### queenofbabes

Try using Gauss's Law? =)

9. Aug 24, 2009

### Luchopas

no text book just some cliff notes i founded,
here i uploaded my argument for the electrix field outside the sphere

sorry this one it is:

http://img24.imageshack.us/img24/8111/newpuw.png [Broken]

for r>R

for r<R = ???

Last edited by a moderator: May 4, 2017
10. Aug 24, 2009

### queenofbabes

As stated, you will need Gauss's Law. Go wiki it. You ought to get a text though, cliff notes are horrible for learning.

11. Aug 24, 2009

### Luchopas

ok but how do i proceed with this problem...
i just wnat to know the steps..

12. Aug 24, 2009

### gabbagabbahey

for r<R, the only thing that changes in your calculations is that the charge enclosed by your gaussian surface will no longer be $\frac{4}{3}\pi R^3\rho$, but just some fraction of it....do you know how to calculate that charge?

Last edited by a moderator: May 4, 2017
13. Aug 24, 2009

### Luchopas

yes integrating the charge Q form 0 to R right?
and replacing this result in Q in the gauss equation.

Last edited: Aug 24, 2009
14. Aug 24, 2009

### gabbagabbahey

No, integrate the charge density, $\rho$ over the volume enclosed by your gaussian surface.

$$Q_{\text{enclosed}}=\int \rho dV$$

In this case, $\rho$ is uniform/constant so the integration should be very easy....what do you get?

15. Aug 24, 2009

### Luchopas

THIS:

http://img269.imageshack.us/img269/1200/bew222.png" [Broken]

HOPE IS RIGHT

Last edited by a moderator: May 4, 2017
16. Aug 24, 2009

### gabbagabbahey

Why is there an $R$ in your expression? Isn't $R$ the radius of the entire sphere of charge? When r<R, don't you want to use the radius of your Gaussian surface,$r$ instead?

Last edited by a moderator: May 4, 2017
17. Aug 24, 2009

### Luchopas

sorry you're right then:

http://img145.imageshack.us/img145/9430/wwwwiy.png" [Broken]

Last edited by a moderator: May 4, 2017
18. Aug 24, 2009

### gabbagabbahey

Okay, so what does that make $\textbf{E}$ for r<R?

19. Aug 24, 2009

### Luchopas

this:

http://img339.imageshack.us/img339/6308/53230410.png" [Broken]

so what is the next step?

Last edited by a moderator: May 4, 2017
20. Aug 24, 2009

### gabbagabbahey

Right, so if the vector from the center of a sphere of charge density $\rho$ is $\textbf{r}_{+}$ the field inside the sphere will be

$$\frac{\rho\textbf{r}_{+}}{3\epsilon_0}$$

And the field inside a sphere of charge density $-\rho$ will be

$$\frac{-\rho\textbf{r}_{-}}{3\epsilon_0}$$

if the vector from the center of the sphere to the point inside the sphere is $\textbf{r}_{-}$

Right?

So pick a point inside the cavity in your drawing in post#1, and label the vector from the center of the positively charged sphere to that point $\textbf{r}_{+}$, and label the vector from the center of the negatively charged sphere to that point $\textbf{r}_{-}$....what is the field at that point due to just the negatively charged sphere? How about the field due to just the positively charged sphere? So the total field at that point is ?

21. Aug 24, 2009

### Luchopas

Sorry im not getting it...... could you explain more specific... thanks please

22. Aug 24, 2009

### Luchopas

you mean r+ and r- like the direction of the magnetic field?

23. Aug 25, 2009

### gabbagabbahey

No.

In your solution for the field inside a sphere, what do $r$ and $\hat{r}$ represent?

The answer of course is that $r$ represents the distance from the center of the sphere to the point $P$ inside the sphere that you are measuring the field at. And $\hat{r}$ represent a unit vector pointing from the center of the sphere, to the point $P$.

So, $\textbf{r}\equiv r\hat{r}$ represents the vector from the center of the sphere, to the point $P$. Right?

So if you call the vector from the center of the positively charged sphere, to the point $P$, $\textbf{r}_{+}$ the field due to that sphere, at the point $P$ will be

$$\textbf{E}_+=\frac{\rho \textbf{r}_{+}}{3\epsilon_0}$$

Right?

24. Aug 25, 2009

### Luchopas

yes youre right, so what do we do know for he hollow section?
becuase know we have both electric field to a point P of both spheres....

Last edited: Aug 25, 2009
25. Aug 25, 2009

### gabbagabbahey

Well, if you call the vector from the center of the negatively charged sphere to the point $P$, $\textbf{r}_{-}$ then the field at $P$ due to the negative sphere will be

$$\textbf{E}_-=\frac{-\rho \textbf{r}_{-}}{3\epsilon_0}$$

Right?

So now you have the field due to each individual sphere, what does the superposition principle tell you the total field will be?

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