Gauss' law for concentric circles

[gracy]

Homework Statement

A small conducting spherical shell with inner radius a and outer radius b is concentric with a larger conducting spherical shell with inner radius c and outer radius d. The inner shell has total charge -2q and the outer shell has charge +4q.
a) calculate the electric field (magnitude and direction) in termsof q and the distance r from the common center of the 2 shells fori) r<a; ii) a<r<b; iii) b<r<c; iv) c<r<d; v)r>d.
b) what is the total charge on the i) inner surface of the smallshell; ii) outer surface of the small shell; iii) inner surface of the large shell; iv) outer surface of the large shell.

Homework Equations

E.4πr^2=q/ε0

The Attempt at a Solution

[/B]
This is drawn in my textbook
I have following doubts
1)why charge +4q is distributed /splitted in such a manner i.e +2q &+2q and then why one +2q is put in the circle of inner radius c and one +2q is placed in the circle of inner radius of d.
2)what is the difference between the positions p3 and p4 because my textbook states at P3 electric field would have some magnitude but at P4 electric field would be zero because it lies inside the conductor then why P3 is not considered to be inside the conductor?

 

[blue_leaf77]

In the question it's stated that

The inner shell hastotal charge +2q

But the picture shows otherwise, which is the correct one?

 

[gracy]

But the picture shows otherwise, which is the correct one?

I have fixed that.

 

[STEMucator]

1)why charge +4q is distributed /splitted in such a manner i.e +2q &+2q and then why one +2q is put in the circle of inner radius c and one +2q is placed in the circle of inner radius of d.

Let me start by saying the innermost cavity with radius $a$ is going to have zero enclosed charged, and therefore the electric field is zero inside of the cavity.

Now, what about between $a$ and $b$? Remember the charge is distributing itself uniformly across the outer surface of the inner conductor.

Then the diagram indicates a $-2 q$ charge on the outside of the innermost shell and a $+2 q$ charge on the inside of the outermost shell. This indicates an electric field that points from the inside of the outermost shell to the outside of the innermost shell. You may use Gauss' law to find $\vec E$.

Now what is the electric field between $c$ and $d$? Hint: Think about the field between $a$ and $b$.

Finally, the outermost shell has a $+2 q$ charge on its surface, and therefore the electric field points radially outward from the sphere. Use Gauss' law to find the electric field once again.

2)what is the difference between the positions p3 and p4 because my textbook states at P3 electric field would have some magnitude but at P4 electric field would be zero because it lies inside the conductor then why P3 is not considered to be inside the conductor?

Well for one, the two points are at different locations. The electric field at $P3$ is due to the opposing charges on the inside of the outer conductor and outside of the inner conductor. There must be an electric field at $P3$ as a result, whereas at $P4$ the electric field should be zero because you are on the conductor itself. $P3$ is not inside a conductor because it is outside the inner conductor which has a charge on its surface and $P3$ is inside the outer conductor which also has a charge on its inner surface.

 

[gracy]

P1 is also not inside the conductor,right?

 

[gracy]

This question is still remaining.

1)why charge +4q is distributed /splitted in such a manner i.e +2q &+2q and then why one +2q is put in the circle of inner radius c and one +2q is placed in the circle of inner radius of d.

 

[blue_leaf77]

1)why charge +4q is distributed /splitted in such a manner i.e +2q &+2q and then why one +2q is put in the circle of inner radius c and one +2q is placed in the circle of inner radius of d.

Such charge redistribution for the larger sphere is required to satisfy the requirement that the electric field inside a conductor must be zero. In order to have $E=0$ for $c<r<d$, the net charge enclosed by a spherical surface with radius $r$ with $c<r<d$ must be zero. Since you already have $-2q$ total charge from the small sphere, you need $+2q$ to be placed on the inner surface of the large sphere. You are left with the remaining $+2q$ in the large sphere and these charges must reside on the outer surface for the same reason as above, $E=0$ inside any conductor.

 

[Doc Al]

1)why charge +4q is distributed /splitted in such a manner i.e +2q &+2q and then why one +2q is put in the circle of inner radius c and one +2q is placed in the circle of inner radius of d.

Imagine a Gaussian surface inside (within the material) the outer conductor (intersecting P4). What's the field over that Gaussian surface? Apply Gauss' law.

 

[gracy]

What's the field over that Gaussian surface?

zero.electric field inside the conductor (here gaussian surface) is zero.

 

[Doc Al]

zero.electric field inside the conductor (here gaussian surface) is zero.

Good. So what must the total charge enclosed by that surface be?

 

[gracy]

So what must the total charge enclosed by that surface be?

zero.

 

[Doc Al]

zero.

Good. So what's the total charge on the inner conductor? Given that, what must be the charge on the inner surface of the outer conductor to cancel it?

 

[gracy]

So what's the total charge on the inner conductor? Given that, what must be the charge on the inner surface of the outer conductor to cancel it?

the net charge enclosed by a spherical surface with radius rr with c<r<dc must be zero. Since you already have −2qtotal charge from the small sphere, you need +2q to be placed on the inner surface of the large sphere. You are left with the remaining +2q in the large sphere and these charges must reside on the outer surface for the same reason as above, E=0 inside any conductor.

Got your points.Thanks @Doc Al & @blue_leaf77

 

[gracy]

The same question asks me to make a graph of E as a function of r.Could you please help me in that too?

 

[blue_leaf77]

Have you calculated the electric field in each region?

 

[gracy]

Have you calculated the electric field in each region?

Yes.

 

[blue_leaf77]

Then, making a graph of E(r) should not be a problem.

 

[gracy]

Hmm...let's see.

 

[gracy]

I want to know how the blue lines are figured out?

 

[blue_leaf77]

I think the question asks you to calculate the magnitude of the field, and hence should be positive everywhere.
Figure out the blue lines, you mean the values on the vertical axis? You have figured it out, haven't you?

 

[gracy]

You have figured it out, haven't you?

It is (the graph)given as solution in backside of my book.

Have you calculated the electric field in each region?

That I have calculated on my own but only for the below cases

i) r<a; ii) a<r<b; iii) b<r<c; iv) c<r<d; v)r>d.

 

[gracy]

And I don't understand where these values are coming from
1/4πε0.2qd^2,1/4πε0.-2q c^2,1/4πε0.-2qb^2

 

[blue_leaf77]

Let me ask you this, what expression did you get for the E field for $b<r<c$?

 

[STEMucator]

The enclosed charge itself between $b$ and $c$ is?

 

[gracy]

what expression did you get for the E field for b<r<cb

you mean for P3?It is E=-2q/4πε0r^2

 

[gracy]

The enclosed charge itself between bb and cc is?

zero.

 

[blue_leaf77]

you mean for P3?It is E=-2q/4πε0r^2

Yes. So what will you get after setting r=b and r=c?

 

[gracy]

Yes. So what will you get after setting r=b and r=c?

But for b &c E is zero.

 

[blue_leaf77]

But for b &c E is zero.

This is an example of boundary condition for the normal components of electric field. Mathematically, it has the form of
$$
E_{1\perp}-E_{2\perp} = \frac{\sigma}{\epsilon_0}
$$
where there is surface charge density, the normal components of the E field in the media separated by the surface undergo discontinuity.

 

[gracy]

This is an example of boundary condition for the normal components of electric field. Mathematically, it has the form of

E1⊥−E2⊥=σϵ0​

E_{1\perp}-E_{2\perp} = \frac{\sigma}{\epsilon_0}
where there is surface charge density, the normal components of the E field in the media separated by the surface undergo discontinuity.

All these new for me;did not understand.

 

[blue_leaf77]

That equation physically means that if in the interface between two media there is surface charge density $\sigma$, the electric field components perpendicular to this interface in the two media separated are discontinuous by the amount $\sigma/\epsilon_0$. Let's take an example for the spherical surface of radius $b$ in your problem. This surface separates the conducting medium below ($r<b$) and the vacuum outside ($r>b$). This spherical surface also carries a surface charge density of
$$
\sigma = \frac{-2q}{Area\hspace{1mm}of\hspace{1mm} the\hspace{1mm} surface} = \frac{-2q}{4\pi b^2}
$$
We know that the field in immediately below the surface is zero since it's inside the a conductor, so $E_{1\perp}=0$. The perpendicular component of the field immediately outside our spherical surface is then
$$
E_{2\perp} = E_{1\perp} - \frac{\sigma}{\epsilon_0} = 0 - \frac{-2q}{4\pi \epsilon_0 b^2} = \frac{2q}{4\pi \epsilon_0 b^2}
$$
But since the E field vector itself in our problem is perpendicular on our spherical surface, we have that $E(b) = E_{2\perp} = \frac{2q}{4\pi \epsilon_0 b^2}$.

 

[gracy]

E2⊥=2q4πϵ0b2E(b) = E_{2\perp} = \frac{2q}{4\pi \epsilon_0 b^2}.

But It is (negative) -2q/4πϵ0b^2 in the graph.

 

[BvU]

So it's a matter of interpreting blue leaf's 1 and 2 in post #29 and in post #31, right ?
My guess is (and this guy backs me up in his (634) ) he mixed them up. If you go from 1 to 2 over a surface charge $\sigma$ you have $E_{2\perp}-E_{1\perp} = \frac{\sigma}{\epsilon_0}$.

This way the field in $b < r < c$ is the field of -2q at the origin -- as it should be.

Re your

But for b &c E is zero.

that is true coming from the conductor side. In his post #27 he wants you to come from the other side, so you get a value for the jump in E at the discontinuity.

 

[gracy]

But for b &c E is zero

I think I am confusing the point P2 with b.Electric field at r=b was never zero.

 

[gracy]

And sign confusion is still there .Why for r=d it's positive and for r=b &r=c it's negative.What pattern does it follow?from outside to inside or from inside to outside?

 

[Doc Al]

And sign confusion is still there .Why for r=d it's positive and for r=b &r=c it's negative.What pattern does it follow?from outside to inside or from inside to outside?

Consider the sign of the charge enclosed by Gaussian spheres at those points.

 

[gracy]

Consider the sign of the charge enclosed by Gaussian spheres at those points.

But it also depends on

E2⊥−E1⊥

E1⊥−E2⊥

 

[Doc Al]

But it also depends on

That's an alternate way of looking at it. Mine is easier.

 

[gracy]

That's an alternate way of looking at it

So which formula are you using?

 

[Doc Al]

So which formula are you using?

I don't need a 'formula' to tell me the direction of the electric field! If the enclosed charge is positive, the field points outward; if negative, inward.

 

[gracy]

I don't need a 'formula' to tell me the direction of the electric field

E2⊥−E1 or E1⊥−E2

 

[blue_leaf77]

DocAl's way is actually more practical. But if you insist on knowing how to really decide between above - below or the other way around, I will go into a slightly deeper math, which I hope you can still grasp. See the picture I attached. The gray plane surface is actually a multiple zoom in result of a certain portion of your spherical conducting surface. Make a cylindrical volume such that part of it lies above the conductor, other part lies below it. We have Gauss law $\oint \mathbf{E}\cdot d\mathbf{a} = Q_{enc}/\epsilon_0 = \int \mathbf{E}_{above}\cdot d\mathbf{a} + \int \mathbf{E}_{below}\cdot d\mathbf{a}$, the integral over the side surface of the cylinder should be zero. To answer your confusion, we actually really have to know the direction of the fields in each region. So, the minus sign in $E_{2\perp} - E_{1\perp}$ can even be a positive sign, depending on the relative directions of the fields in the two regions. The source I was using when I brought up that equation for the first time assumes that the field in both regions point in the same direction, so the minus sign appears. As a conclusion, to determine the correct sign you need to know the direction of the fields in each region. In your problem, the charge on surface r=b is negative, so the fields outside it must point inward.

 

[gracy]

See the picture I attached

where is it?

 

[gracy]

DocAl's way is actually more practical.

But I did not understand his way.

 

[gracy]

I think I am confusing the point P2 with b.Electric field at r=b was never zero.

But for b &c E is zero.

right?

 

[gracy]

E1⊥

it does not have any magnitude so no direction.And E2⊥ will have negative sign.

 

[gracy]

If the enclosed charge is positive, the field points outward; if negative, inward.

if no charges are enclosed?how am I going to decide direction?

 

[blue_leaf77]

if no charges are enclosed?how am I going to decide direction?

If there is no surface charge density, the normal field components are continuous.

 

[Doc Al]

if no charges are enclosed?how am I going to decide direction?

If no charges are enclosed within your Gaussian sphere (and there is symmetry), then what will the field at the surface be?

 

[gracy]

then what will the field at the surface be?

Zero.