# Electric Field & Potential for Three Concentric spheres

• Sudharshan
In summary: I am not sure if it is completely correct. I would greatly appreciate it if someone could take a look at it and tell me if it is correct.
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.

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 ?

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?

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.

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.]

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.

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.

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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

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.

Tanya Sharma said:
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 ?

Tanya Sharma said:
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.

TSny said:
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 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?

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##.

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

Yes. That looks correct.

That all looks good to me.

Ok. Awesome. Thanks for all your help!

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 .

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Tanya Sharma said:
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}##.

TSny said:
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 .

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.

1 person
TSny said:
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

Tanya Sharma said:
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 .

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.

Tanya Sharma said:

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.

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 ?

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}##.

1 person
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 ?

Tanya Sharma said:
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:

## What is an electric field?

An electric field is a vector quantity that describes the strength and direction of the force exerted on a charged particle by other charged particles in its vicinity. It is represented by arrows that point in the direction of the force on a positive charge and the length of the arrows indicate the strength of the field.

## How is an electric field calculated for three concentric spheres?

The electric field for three concentric spheres can be calculated using the principle of superposition, which states that the total electric field at a point is equal to the sum of the individual electric fields from each charge. The formula for calculating the electric field is E = kq/r^2, where k is the Coulomb constant, q is the charge, and r is the distance from the charge to the point of interest.

## What is the potential difference between two points in an electric field?

The potential difference between two points in an electric field is a measure of the work done per unit charge in moving a test charge from one point to another. It is also known as voltage and is measured in volts (V). The formula for calculating potential difference is V = Ed, where E is the electric field and d is the distance between the two points.

## How does the potential vary between the three concentric spheres?

In a system of three concentric spheres, the potential decreases as you move from the outer sphere to the inner sphere. This is because the electric potential at a point is directly proportional to the amount of charge and inversely proportional to the distance from the source of the charge. Therefore, the potential decreases as you move closer to the source of the charge.

## How does the electric field change with the change in charge or distance in three concentric spheres?

The electric field in a system of three concentric spheres changes with the change in charge or distance according to the inverse-square law. This means that the electric field is directly proportional to the magnitude of charge and inversely proportional to the square of the distance from the charge. Therefore, a change in either the charge or distance will result in a change in the electric field.

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