EM field between 2 superimpose spheres

[Luchopas]

Homework Statement

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"

Homework Equations

Find the electric field in the hollow section.

The Attempt at a Solution

?'

 

[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?

 

[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.

 

[Luchopas]

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?

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...

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

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...

 

[gabbagabbahey]

for a solid sphere the elctric field inside and outside is

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

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?

 

[Luchopas]

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?

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 I am new in this physics subject

 

[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?

 

[queenofbabes]

Try using Gauss's Law? =)

 

[Luchopas]

no textbook 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

for r>R

for r<R = ?

 

[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.

 

[Luchopas]

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

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

 

[gabbagabbahey]

no textbook 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

for r>R

for r<R = ?

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?

 

[Luchopas]

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?

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

 

[gabbagabbahey]

yes integrating the charge Q form 0 to R right?

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?

 

[Luchopas]

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?

THIS:

http://img269.imageshack.us/img269/1200/bew222.png"

HOPE IS RIGHT

 

[gabbagabbahey]

THIS:

http://img269.imageshack.us/img269/1200/bew222.png"

HOPE IS RIGHT

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?

 

[Luchopas]

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?

sorry you're right then:

http://img145.imageshack.us/img145/9430/wwwwiy.png"

 

[gabbagabbahey]

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

 

[Luchopas]

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

this:

http://img339.imageshack.us/img339/6308/53230410.png"

so what is the next step?

 

[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 ?

 

[Luchopas]

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 ?

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

 

[Luchopas]

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

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

 

[gabbagabbahey]

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

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?

 

[Luchopas]

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?

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...

 

[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?

 

[Luchopas]

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?

It will be the sum of both electric fields ...right?

 

[gabbagabbahey]

Right, it will be the vector sum of the two fields...which will be?

 

[Luchopas]

Right, it will be the vector sum of the two fields...which will be?

(ro/3*epsilon 0)* (r+ - r-)

right?

 

[gabbagabbahey]

Right,

$$\textbf{E}=\frac{\rho}{3\epsilon_0}\left(\textbf{r}_+ - \textbf{r}_- \right)$$

But, now look at your original diagram and draw in the vectors $\textbf{r}_+$ and $\textbf{r}_{-}$...what is the vector $\textbf{r}_{+}-\textbf{r}_{-}$ equal to?

 

[Luchopas]

Right,

$$\textbf{E}=\frac{\rho}{3\epsilon_0}\left(\textbf{r}_+ - \textbf{r}_- \right)$$

But, now look at your original diagram and draw in the vectors $\textbf{r}_+$ and $\textbf{r}_{-}$...what is the vector $\textbf{r}_{+}-\textbf{r}_{-}$ equal to?

it is equal to the d vector ..right?

 

[gabbagabbahey]

Right!

So, $$\textbf{E}=\frac{\rho\textbf{d}}{3\epsilon_0}$$

 

[Luchopas]

Right!

So, $$\textbf{E}=\frac{\rho\textbf{d}}{3\epsilon_0}$$

got it, thank you very much.

 

[gabbagabbahey]

Well, it's your homework problem so it doesn't really matter whether I'm sure, what's important is whether or not you're sure.

You seem to have doubts... is there a reason you doubt this result? Is there some part of the above analysis you are uncomfortable with?

 

[Luchopas]

Well, it's your homework problem so it doesn't really matter whether I'm sure, what's important is whether or not you're sure.

You seem to have doubts... is there a reason you doubt this result? Is there some part of the above analysis you are uncomfortable with?

mm not really just was a moment of doubt, but you got me convinced.
thanks