Electric Field & Potential for Three Concentric spheres

[Sudharshan]

Homework Statement

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

Homework Equations

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

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.

 

[Tanya Sharma]

Hi Sudharshan

On Page 2 while applying Gauss law to a spherical surface of radius r (b<r<c) and calculating Q_free , 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 ?

 

[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?

 

[Tanya Sharma]

I think you need to take algebraic sum of the two charges i.e ρ₁V₁+ρ₂V₂ , where ρ is the density and V is the volume of respective regions.

 

[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.]

 

[Sudharshan]

Hey,

I have now done this correction that you pointed out. I can't believe it was quite this easy (if i have understood what you meant correctly). I have linked my solution for E(r) for all r now. Is it correct?

Page 1: http://i.imgur.com/hiEvBPw.jpg
Page 2: http://i.imgur.com/0mdblxs.jpg
Page 3: http://i.imgur.com/dFcwdPg.jpg

 

[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.

 

[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.

 

[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

 

[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.

 

[Sudharshan]

Hi Sudharshan

On Page 2 while applying Gauss law to a spherical surface of radius r (b<r<c) and calculating Q_free , 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 ?

I think you need to take algebraic sum of the two charges i.e ρ₁V₁+ρ₂V₂ , where ρ is the density and V is the volume of respective regions.

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.]

As stated by Tanya, I take the algebraic sum of the two charges i.e ρ₁V₁+ρ₂V₂ 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?

 

[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 $\varepsilon_2$.

 

[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

 

[TSny]

Yes. That looks correct.

 

[Sudharshan]

I finally managed to calculate the V(r) for all r by using the E(r) for all r that I had calculated. It would be great if someone could take a look at it to see if it is correct. Thank you.

Page 1: http://i.imgur.com/fKsv7j4.jpg
Page 2: http://i.imgur.com/uCL7amL.jpg
Page 3: http://i.imgur.com/EtkbYNa.jpg
Page 4: http://i.imgur.com/XVbPfw8.jpg
Page 5: http://i.imgur.com/wS7HqDc.jpg

 

[TSny]

That all looks good to me.

 

[Sudharshan]

Ok. Awesome. Thanks for all your help!

 

[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 .

Please give your suggestions .

 

[TSny]

There is something which confuses me every now and then .

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

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.

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

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

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), $\vec{dl} = \vec{dr} = dr\hat{r}$.

 

[Tanya Sharma]

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), $\vec{dl} = \vec{dr} = dr\hat{r}$.

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 .

 

[TSny]

I think $\vec{dl}$ and $\vec{dr}$ are different notations for the same quantity: an infinitesimal displacement. If you want the potential difference in going from point a to point b, then
$ΔV = V_b - V_a = -\int_{a}^{b} \vec{E} \cdot \vec{dl} = -\int_{a}^{b} \vec{E} \cdot \vec{dr}$ for any path between a and b.

$\vec{dl}$ and $\vec{dr}$ denote an infinitesimal displacement along the path.

Setting $a$ = some point at ∞ where V = 0,

$V(r) = -\int_{∞}^{r} \vec{E} \cdot \vec{dl} = -\int_{∞}^{r} \vec{E} \cdot \vec{dr}$ for any path.

If you choose a radial path, then $\vec{dr} = dr \hat{r}$. As you come in radially from infinity, $\vec{dr}$ should point radially inward. That agrees with $dr \hat{r}$ which also point radially inward since $dr$ is negative as you move radially inward and $\hat{r}$ always points radially outward.

Not sure if this is helping.

 

[Tanya Sharma]

If you choose a radial path, then $\vec{dr} = dr \hat{r}$. As you come in radially from infinity, $\vec{dr}$ should point radially inward. That agrees with $dr \hat{r}$ which also point radially inward since $dr$ is negative as you move radially inward and $\hat{r}$ always points radially outward.

Here lies the problem

This is exactly what I asked you in post #18

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 .

Please check my understanding .

1. $\vec{dr}$ can point either radially outwards or inwards , depending on whether we are moving from (r→∞) or (∞→r)

2. In both the cases $\vec{dl} = \vec{dl} = dr \hat{r}$ .

3. When we are moving from ∞→r , $\vec{dr}$ points inwards and dr is negative ,whereas when we are moving from r→∞ , $\vec{dr}$ points outwards and dr is positive.

 

[TSny]

Please check my understanding .

1. $\vec{dr}$ can point either radially outwards or inwards , depending on whether we are moving from (r→∞) or (∞→r)

2. In both the cases $\vec{dl} = \vec{dl} = dr \hat{r}$ .

3. When we are moving from ∞→r , $\vec{dr}$ points inwards and dr is negative ,whereas when we are moving from r→∞ , $\vec{dr}$ points outwards and dr is positive.

That's my understanding. $\vec{dr}$ points in whatever direction you are moving along the path of integration.

 

[Tanya Sharma]

I think what you are referring as $\vec {dr}$ ,I am referring as $\vec {dl}$ .$\vec {dl}$ is displacement of the test charge (in this case I am treating it radial).But it can be along any path .My use of $\vec {dr}$ is the displacement along the radial direction . So, in a sense $\vec {dr}$ is the radial component of $\vec {dl}$ .The other component of $\vec {dl}$(tangential) doesn't contribute to the work done , being at 90° to the electric field .

Do I make sense ?

 

[TSny]

I guess you can define $\vec{dr}$ that way. That is, you define it as $dr\hat{r}$ where $dr$ is the change in r for some displacement $\vec{dl}$ that is not necessarily radial. I'm not sure I see any advantage of that definition.

The common usage of $\vec{dr}$ is to denote an infinitesimal dispacement in any direction (not necessarily in the radial direction.) You find many introductory textbooks that denote $\vec{r}$ for position of a particle and $\Delta \vec{r}$ for a change in position, or displacement, in any direction (not necessarily radial). $\vec{dr}$ is just the infinitesimal version of $\Delta\vec{r}$.

 

[Tanya Sharma]

Okay...I will have to rethink on this.I have been harbouring some misconceptions for a long time .

Thank you very much for your time and elegant explanations . I am glad I put this question in front of you .

Should I say Good Night to you ?

 

[TSny]

Okay...I will have to rethink on this.I have been harbouring some misconceptions for a long time .

Thank you very much for your time and elegant explanations . I am glad I put this question in front of you .

Should I say Good Night to you ?

It is that time...:zzz: