Point Charge in a Glass Sphere

In summary: Keep it up!In summary, the problem involves a point charge inside a glass sphere with a dielectric constant of 10, and this sphere is enclosed in a hollow metal sphere with an inner radius of 2 m and a thickness of 0.5 m. The task is to find the electric field strength in various regions, including inside the glass sphere, in the gap between the glass and metal, in the metal layer, and on the outside. Using Gauss' law of electrostatics, it can be determined that the field outside of the metal sphere is equal to the field of a point charge (E(r) = Q/(4πε0r2)). The field inside the glass sphere can be calculated using the equation E
  • #1
Lunar_Lander
38
0

Homework Statement



There is a point charge Q = 4π nC inside a glass sphere which has the radius 1 m and the dielectric constant εr=10. This sphere is enclosed in a hollow metal sphere with an inner radius of 2 m and a thickness of 0.5 m.

The task is to find the amount of the electric field strength in the glass sphere, in the gap between glass and metal, in the metal layer and on the outside.

Homework Equations


Gauss law of electrostatics.

The Attempt at a Solution


As far as I know, the field outside of the metal sphere should be as if the sphere would be a point charge, i.e.E(r)=Q/(4πε0r2). But I am a bit lost at the other parts. Is the field in the glass sphere for instance simply: E(r)=Q/(4π εrε0r2)?
 
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  • #2
Lunar_Lander said:

Homework Statement



There is a point charge Q = 4π nC inside a glass sphere which has the radius 1 m and the dielectric constant εr=10. This sphere is enclosed in a hollow metal sphere with an inner radius of 2 m and a thickness of 0.5 m.

The task is to find the amount of the electric field strength in the glass sphere, in the gap between glass and metal, in the metal layer and on the outside.

Homework Equations


Gauss law of electrostatics.

The Attempt at a Solution


As far as I know, the field outside of the metal sphere should be as if the sphere would be a point charge, i.e.E(r)=Q/(4πε0r2).
Correct. :approve: (Of course this is only correct due to the spherical symmetry of the whole thing. If it were any other shape, the answer would be different.)
But I am a bit lost at the other parts. Is the field in the glass sphere for instance simply: E(r)=Q/(4π εrε0r2)?
Yes, the equation for inside the glass sphere is also correct. :approve:

You should be able to use the standard version of Gauss' law for the region (gap) between the glass sphere and hollow metal shell (εr = 1 in this region).

The electric field within a conductor is special [Edit: by that I mean within the conducting material itself]. But I'll leave that for you to look up. :smile:
 
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  • #3
OK, thanks :)! In the metal, there should be a shift of charges due to the field from the inside, so that the inside of the metal layer is free of the electric field, i.e. |E(r)|=0?

One last pointer on the fields inside though, please. Do I need to incoporate the radius of the glass sphere into the calculation somehow? Or can I just say that below r0 (which is the radius of the glass sphere, the inner radius of the metal was denoted as r1 and the thickness was given as d, so the outer surface of the metal sphere is at r1+d), the field is E(r<r0)=Q/(4π*ε0r*r2), then in the air gap, the field is E(r0<r<r1)=Q/(4π*ε0*r2)?

And accordingly, E(r>r1+d)=Q/(4π*ε0*r2)?
 
  • #4
Lunar_Lander said:
OK, thanks :)! In the metal, there should be a shift of charges due to the field from the inside, so that the inside of the metal layer is free of the electric field, i.e. |E(r)|=0?
Yes that's right. The (static) electric field within a conductor is always zero. Things get more complicated when you move on to electrodynamics. But with static charges, the field within a conducting material is always zero. :approve:
One last pointer on the fields inside though, please. Do I need to incoporate the radius of the glass sphere into the calculation somehow? Or can I just say that below r0 (which is the radius of the glass sphere, the inner radius of the metal was denoted as r1 and the thickness was given as d, so the outer surface of the metal sphere is at r1+d), the field is E(r<r0)=Q/(4π*ε0r*r2),
The latter looks good to me. :approve:
then in the air gap, the field is E(r0<r<r1)=Q/(4π*ε0*r2)?
Yes, that's right! :approve: And that's for the same reason that you can say E = Q/(4π*ε0*r2) outside of the metal shell too. As long as there is perfect spherical symmetry, it doesn't matter what is on the outside of the Gaussian surface. All that matters is the total charge within.
And accordingly, E(r>r1+d)=Q/(4π*ε0*r2)?
Yes. :approve: Good work.
 
  • #5


Thank you for your question. I would approach this problem by first understanding the basic principles involved. In this case, we are dealing with electrostatics, which is the study of electric charges at rest. The key equation we will use is Gauss's law, which relates the electric flux through a closed surface to the enclosed charge.

In this problem, we have a point charge Q located inside a glass sphere with a dielectric constant εr = 10. The glass sphere is then enclosed in a hollow metal sphere with an inner radius of 2 m and a thickness of 0.5 m. Our goal is to find the electric field strength at various points in and around the system.

To begin, we can use Gauss's law to find the electric field strength inside the glass sphere. Since the glass sphere is a closed surface, the electric flux through it is equal to the enclosed charge divided by the permittivity of free space (ε0). Therefore, we can write:

∮E⃗⋅dA⃗ = Q/ε0

Where ∮E⃗⋅dA⃗ is the electric flux through the surface of the glass sphere. Since the electric field is constant and perpendicular to the surface of the sphere, we can simplify this equation to:

E(r)4πr2 = Q/ε0

Where E(r) is the electric field strength at a distance r from the point charge. Since we know the value of Q and the radius of the glass sphere, we can solve for E(r):

E(r) = Q/(4πε0r2)

This is the electric field strength inside the glass sphere. Notice how the dielectric constant εr does not appear in this equation. This is because the electric field inside a dielectric material is reduced by a factor of εr compared to the electric field in free space. Therefore, the electric field inside the glass sphere is simply the electric field due to the point charge Q, divided by the dielectric constant εr.

Now, to find the electric field in the gap between the glass and metal spheres, we can use a similar approach. The electric flux through the surface of the gap is equal to the charge enclosed by the glass sphere and the metal sphere, divided by ε0. Therefore, we can write:

∮E⃗⋅dA⃗ = (Q+Q
 

1. What is a point charge in a glass sphere?

A point charge in a glass sphere refers to a hypothetical situation where a single positive or negative charge is concentrated at a single point on the surface of a glass sphere.

2. How is a point charge in a glass sphere different from a regular charge?

A regular charge is spread out over a finite area, while a point charge is treated as a single infinitesimal point. This simplification is often used in theoretical calculations and models.

3. What is the significance of studying a point charge in a glass sphere?

Studying a point charge in a glass sphere can provide insights into the behavior of electric fields and forces in a simplified scenario. It can also serve as a useful theoretical tool in understanding more complex systems.

4. How is the electric field affected by a point charge in a glass sphere?

The electric field around a point charge in a glass sphere is spherically symmetric, meaning it has the same strength and direction at all points on the surface of the sphere. It follows the inverse-square law, decreasing with the square of the distance from the point charge.

5. Can a point charge exist in a glass sphere in real life?

In reality, it is not possible to have a point charge in a glass sphere as all charges are spread out over a finite area. However, the concept of a point charge is useful in theoretical calculations and can be used to approximate real-world scenarios.

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