# Electric Field & Potential for Three Concentric spheres

1. Feb 8, 2014

### Sudharshan

1. The problem statement, all variables and given/known data
Three volumes bounded by three concentric spheres with radii a, b ​​and c. The innermost volume r<a, consists of vacuum. Next volume, a<r<b, is filled with a material having a constant volume charge density ρ1 and a relative dielectric constant ε1. The external volume, b<r<c, consists of a different material with ρ2 respective ε2. Outside the spheres, r> c, in vacuum.

Find E(r), V(r) for all r.
I.e. r<a, a<r<b, b<r<c, r>c

2. Relevant equations

Guass's Law, http://en.wikipedia.org/wiki/Gauss%27s_law
Electric Potential, http://en.wikipedia.org/wiki/Electric_potential

3. The attempt at a solution
I have calculated the E(r) and V(r) for all r. However, I am not sure about certain E(r), especially E(r) for b<r<c and r>c which leads to my total E(r) and the calculations of V(r) of possibly being wrong. I am also not certain about certain V(r), especially V(r) for b<=r<=c and r<=a. Is my solution correct? Is E(r) and V(r) for all r correct?

Page 1: http://i.imgur.com/XckQoCM.jpg Equations used for calculation of E(r), Calculation of E(r)
Page 2: http://i.imgur.com/5ldfQHL.jpg Calculation of E(r)
Page 3: http://i.imgur.com/bn7ui8S.jpg Calculation of E(r), Total E(r) for all r.
Page 4: http://i.imgur.com/1SxJAP6.jpg Equation used for calculation of V(r), Calculation of V(r)
Page 5: http://i.imgur.com/RRCKCOb.jpg Calculation of V(r)
Page 6: http://i.imgur.com/Ihud14e.jpg Calculation of V(r), Final V(r).

Any help is appreciated.

2. Feb 9, 2014

### Tanya Sharma

Hi Sudharshan

On Page 2 while applying Gauss law to a spherical surface of radius r (b<r<c) and calculating Qfree , you have considered charge enclosed in the region b<r .What about charge in the region a<r<b ? Doesn't that come into account ?

Similarly on Page 3,for r>c , you have considered charge enclosed in the region b<r<c .What about charge in the region a<r<b ?

3. Feb 9, 2014

### Sudharshan

Hello Tanya,

These two parts you mentioned are the ones that I am not sure how to calculate. I do not know how to add the charge for region a<r<b for Page 2 and b<r<c for Page 3. Do I just multiply it in ? How do I proceed here?

4. Feb 9, 2014

### Tanya Sharma

I think you need to take algebraic sum of the two charges i.e ρ1V12V2 , where ρ is the density and V is the volume of respective regions.

5. Feb 9, 2014

### TSny

In calculating the charge enclosed by the surface for finding D in the regions b< r< c and r>c, did you include the charge due to ρ1? [EDIT: Sorry, I see Tanya has already pointed this out.]

6. Feb 10, 2014

### Sudharshan

Last edited: Feb 10, 2014
7. Feb 10, 2014

### TSny

That looks almost correct. The only thing I see wrong is that you used incorrect dielectric constants for the regions b<r<c and r>c.

8. Feb 10, 2014

### Sudharshan

Oh yes. I see it now. Forgot to change the constant for b<r<c and remove it (since it is equal to 1 in a vacuum) for region r>c.

Last edited: Feb 10, 2014
9. Feb 10, 2014

### Sudharshan

I have calculated the V(r) for all r by using the E(r) for all r that I had calculated. I am not sure if it is completely correct. It would be great if someone could take a look at it to see if it is correct. I know it is a lot of pages but it would be help me a lot to know if I have solved this question correctly or not. Thank you.

Page 1: http://i.imgur.com/pp56cfs.jpg
Page 2: http://i.imgur.com/OnwIofm.jpg
Page 3: http://i.imgur.com/yZUWYxX.jpg
Page 4: http://i.imgur.com/UmvvpiV.jpg
Page 5: http://i.imgur.com/w6oFNuO.jpg

10. Feb 10, 2014

### TSny

It appears to me that when integrating to find V, you still don't have all of the dielectric constants correct. For example, in region r > c, you still have an $\varepsilon_1$ appearing. In region b<r<c where you integrate from c to r, I see both $\varepsilon_1$ and $\varepsilon_2$ appearing.

11. Feb 10, 2014

### Sudharshan

As stated by Tanya, I take the algebraic sum of the two charges i.e ρ1V12V2 when integrating to find E. Thus for r > c, I have ρ1V1 and p2V2 in the equation which leads to me having ε1 and ε2 in the final equation for E for r > c. This equation is what I use when integrating for V.

Here is my final E(r) for all r (after corrections):
http://i.imgur.com/a3jZ0Jg.jpg

Do you think my E(r) is wrong? If so, what am I doing wrong?

12. Feb 10, 2014

### TSny

You first find D everywhere using Gauss' law. Just make sure that you include all the charge within the Gaussian surface. I think you have done that.

Once you have D in a certain region, then E is found from D using the dielectric constant for that region.

For example, the only dielectric constant that will appear in the electric field expression in the region b < r < c will be $\varepsilon_2$.

13. Feb 11, 2014

### Sudharshan

Ahhhh. I understand now. I see my mistake. I have recalculated with the proper dielectric constants now I hope. Is my E(r) for all r correct now ?

http://i.imgur.com/bI87guV.jpg

14. Feb 11, 2014

### TSny

Yes. That looks correct.

15. Feb 11, 2014

### Sudharshan

16. Feb 11, 2014

### TSny

That all looks good to me.

17. Feb 11, 2014

### Sudharshan

Ok. Awesome. Thanks for all your help!!

18. Feb 11, 2014

### Tanya Sharma

Hi TSny

Thanks for nicely guiding the OP...

There is something which confuses me every now and then .I would be grateful if you could help me in dispelling this confusion .

The problem is with signs while finding the potential at a point .

Let us consider a point charge at the origin O .Now we need to find the potential at a point P at a distance 'r' from O .The definition says that the potential at a point is the work done in bringing a unit test charge from infinity to the point .

$V(r) = -\int_{∞}^{r} \vec{E} \cdot \vec{dl}$ , where $\vec{dl}$ represents the infinitesimal displacement towards O .

Now , the limits are in terms of radial distance ,i.e distances measured from O , so we convert the displacement $\vec{dl}$ in terms of radial displacement $\vec{dr}$.

$\vec{dl} = -\vec{dr} = -dr\hat{r}$

$V(r) = -\int_{∞}^{r} \vec{E} \cdot (-dr\hat{r})$

$V(r) = -\int_{∞}^{r} \frac{kQ}{r^2}\hat{r} \cdot (-dr\hat{r})$

$V(r) = \int_{∞}^{r} \frac{kQ}{r^2}\hat{r} \cdot dr\hat{r}$

$V(r) = \int_{∞}^{r} \frac{kQ}{r^2}dr\hat{r} \cdot \hat{r}$

$V(r) = \int_{∞}^{r} \frac{kQ}{r^2}dr$

$V(r) = -\frac{kQ}{r}$ , which is not right (it has an additional minus sign)

I guess the problem lies somewhere with $\vec{dr}$ . Does $\vec{dr}$ always points away from the origin .In this case the magnitude of $\vec{dr}$ , dr should be inherently negative because the limits are from larger value ∞ to smaller value r .But when we write $\vec{dl} = -\vec{dr} = -dr\hat{r}$ dr is positive .

Last edited: Feb 11, 2014
19. Feb 11, 2014

### TSny

About 50% of the time I get the sign wrong. Since the sign can have only two values I would be just a well off randomly guessing.

I would say the problem lies here. Usually $\vec{dr}$ represents an arbitrary displacement in space (not necessarily radial). So, $\vec{dr}$ and $\vec{dl}$ denote the same thing.

For the specific case of radial displacement (either outward or inward), $\vec{dl} = \vec{dr} = dr\hat{r}$.

20. Feb 11, 2014

### Tanya Sharma

But the variable of integration should correspond with the limits of integration .The limits of integration (∞ → r) are "distances from the charge at O" , so the variable of integration should also be displacement measured from the origin .How can $\vec{dl} = \vec{dr} = dr\hat{r}$ ?

$\vec{dl}$ is the displacement of the test charge ,as it is moved by an external agent towards O. $\vec{dl}$ and $\vec{dr}$ should have opposite signs .