Gauss' Law & charge inside sphere

In summary, according to the homework, the electric field between the spheres is spherically symmetric and has an equivalent strength of 8 kN/C at 8 cm from the center and 15 kN/C at 17 cm from the center. The total charge on the large hollow sphere is 48 nanoCoulomb.
  • #1
LambertoPierto
7
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1. Homework Statement

11gn3eq.jpg


I need to find the total charge inside the small metal sphere, inside the big metal sphere as well as outside the big metal sphere.

Homework Equations



What confuses me is the electric field vector. Since it's only poiting in one direction it can't originate from a single charge, or a group of charges. What could produce such a field?

The Attempt at a Solution


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I know I'm supposed to use Gauss' law.

For the small sphere. The amount and magnitude of electric field lines entering and leaving is the same. Which indicates that there is no charge inside the small sphere. As for the big sphere, inside and outside I honestly have no clue how to proceed. You should multiply the electric field with the perpedicular area. However if there is only one electrif field line, what area? Once again, so greateful for any idées/nods in the right direction!
 
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  • #2
Hello Lamberto, benvenuto :welcome:
LambertoPierto said:
What confuses me is the electric field vector. Since it's only poiting in one direction it can't originate from a single charge, or a group of charges. What could produce such a field?
That's not much of a relevant equation ! But let me help you a bit: the field between the spheres can only be spherically symmetric(can you explain why ?). They just drew one field vector, but they all point to the center. And at 8 cm from the center they are all 15 kN/C strong.
 
  • #3
BvU said:
Hello Lamberto, benvenuto :welcome:
That's not much of a relevant equation ! But let me help you a bit: the field between the spheres can only be spherically symmetric(can you explain why ?). They just drew one field vector, but they all point to the center. And at 8 cm from the center they are all 15 kN/C strong.

Haha I'm sorry I was in a rush writing my question. So the field is spherically symmetric, due to the fact that the charges repell each other and align at a equal distance from each other, in such a pattern that they create a symmetric field? Could that be?

However with your information I used Gauss' law to calculate the charge inside the small sphere accordingly:
q = E x 4 π (8cm)2 x ε0

And for the large sphere the total charge would be that of the small charge, in addition to the q= E x 4 π (17cm)2 x ε0

(the surface area 4 π (8cm)2 and 4 π (17cm)2 because the electric field is 15 000 N/C at 8cm respective 17cm)

Am I on the right track?
 
  • #4
Yes, I think you are on the right track. Don't rush, however, or you make unnecessary errors, like
LambertoPierto said:
And for the large sphere the total charge would be that of the small charge
you of course mean "the total charge would be minus that of the small sphere"...

LambertoPierto said:
the field is spherically symmetric, due to the fact that the charges repel each other and align at a equal distance from each other
yes, a charge on the small sphere separately (no big hollow sphere present) would give a spherically symmetric charge distribution and a spherically symmetric field, and the large sphere does not introduce any disturbances of that symmetry (because it is concentric -- if it would not be concentric you would have a different E-field and a different, non-uniform charge distribution on the inside of the large shell.)
 
  • #5
BvU said:
you of course mean "the total charge would be minus that of the small sphere"...

Yes of course electric field, it's a vector.

Thank you for the help! I have a better understanding of Gauss' law now!

This dose not concern the excerise but it's something I've been wondering about. Maybe you could help me or perhaps this is inappropriate for this forum. My knowledge about charges, conductors etc are lacking. In this case when the large sphere is positively charged (the electric field lines are pointing out of the sphere). Dose it mean that is has NO negativly charged particles (moving freely around) or that it has a greater amount of positive charges.
 
  • #6
Ah ! That is such a nice question that I am overjoyed you ask it !
What was the total charge on this big hollow sphere again ? 48 nanoCoulomb ?

Calculate with me:
The mass is ##{4\pi\over 3}(0.15^3 - 0.1^3) * \rho \approx 80 ## kg ##\quad (\rho \approx 8000 {\rm \ kg/m^3}) \ ## Pretty heavy !.
With an atomic weight of 55.85 kg/kmol that is 1.4 kmol.
If neutral, each atom counts 26 electrons, so ## 1.4 * 26 * N_A \approx 2.2 \times 10^{25} \ ## electrons in the sphere, a whopping ## 2.2 \times 10^{25} \times q_e \approx 3.6\ ## MegaCoulomb.

So if one in 74 billion electrons is missing you already get this electric field. There are plenty of electrons left over.

I did cheat for effect: most of the Fe electrons are not moving about freely. Only the ones in the conduction band -- but that's a field of physics that I don't know much about. I suppose that on average it's one or two of the outer electrons that move pretty freely.

Please correct my calculation errors -- what was this again about being in a rush :smile: ?
 
  • #7
BvU said:
Ah ! That is such a nice question that I am overjoyed you ask it !
What was the total charge on this big hollow sphere again ? 48 nanoCoulomb ?

Calculate with me:
The mass is ##{4\pi\over 3}(0.15^3 - 0.1^3) * \rho \approx 80 ## kg ##\quad (\rho \approx 8000 {\rm \ kg/m^3}) \ ## Pretty heavy !.
With an atomic weight of 55.85 kg/kmol that is 1.4 kmol.
If neutral, each atom counts 26 electrons, so ## 1.4 * 26 * N_A \approx 2.2 \times 10^{25} \ ## electrons in the sphere, a whopping ## 2.2 \times 10^{25} \times q_e \approx 3.6\ ## MegaCoulomb.

So if one in 74 billion electrons is missing you already get this electric field. There are plenty of electrons left over.

I did cheat for effect: most of the Fe electrons are not moving about freely. Only the ones in the conduction band -- but that's a field of physics that I don't know much about. I suppose that on average it's one or two of the outer electrons that move pretty freely.

Please correct my calculation errors -- what was this again about being in a rush :smile: ?

Dude haha chill, but thank you for your input!

EDIT: sorry not sure how that sounded but I am truly greatful :)
 
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  • #8
BvU said:
Ah ! That is such a nice question that I am overjoyed you ask it !

But hey since you're this enthusiastic! Potential energy in a current. Positive charges loses potential energy as they move towards the negativ output whereas negativ charges gain potential energy as they move towards the positive output. Do the charged not interact with each other as they move through the wire. And what happens with potential energy when it converts to heat or alike due to resistance? Like what happens to charges in a current as they run into a resistor?
 
  • #9
As we have seen, only a very small fraction of the charge that is present in a conductor has to be added or removed to generate a considerable electrical field.

The way I learned it is that in general mobile charge carriers are electrons. But an electron that moves from A to B is completely equivalent to an equal but opposite charge moving from B to A. In transistors and diodes these "missing electrons" are called holes.

The charges interact with each other: they still repel each other so -- as in your charged spheres -- they tend to 'travel' far apart.

Resistance occurs if a material doesn't allow the charge carriers to move completely freely; a fraction of their energy is converted to motion of the consituents of the material (the atoms) and that is heat.
 

Related to Gauss' Law & charge inside sphere

What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed within that surface. It is named after the German mathematician and physicist Carl Friedrich Gauss.

What is the formula for Gauss' Law?

The mathematical expression for Gauss' Law is given by ΦE = Qenc0, where ΦE is the electric flux through a closed surface, Qenc is the charge enclosed within that surface, and ε0 is the permittivity of free space.

How does Gauss' Law apply to a charge inside a sphere?

When a charge is placed inside a conducting sphere, the electric field inside the sphere is constant and directed towards the center. This is because the charge on the surface of the conductor redistributes itself in such a way that the electric field inside cancels out, according to Gauss' Law.

What is the electric field inside a conducting sphere with a charge at its center?

The electric field inside a conducting sphere with a charge at its center is given by E = Q/(4πε0R2), where Q is the charge at the center and R is the radius of the sphere. This expression is valid for both positive and negative charges.

What happens to the electric field inside a conducting sphere if the charge is not at its center?

If the charge is not at the center of the conducting sphere, the electric field inside will still be constant and directed towards the center, but its magnitude will vary with distance from the charge. This is because the charge distribution on the surface of the conductor will be asymmetrical, leading to an asymmetrical cancellation of the electric field inside, as described by Gauss' Law.

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