Gauss' Law with a Insulating Shell

In summary, a spherical shell insulator with inner radius a = 4 cm and outer radius b = 6 cm carries a total charge of Q = + 9 μC and the task is to find the y-component of the electric field at point P which is located at (x,y) = (0, -5 cm). Using the equation E∮dA = Qenc/ε0, the solution involves analyzing the problem with a Gaussian surface that ends at point P and finding the volume of the spherical shell. After finding the volume, the charge density can be calculated using the equation ρ = Q/(4/3π(b^3-a^3)).
  • #1
PenDraconis
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Homework Statement


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An insulator in the shape of a spherical shell is shown in cross-section above. The insulator is defined by an inner radius a = 4 cm and an outer radius b = 6 cm and carries a total charge of Q = + 9 μC(You may assume that the charge is distributed uniformly throughout the volume of the insulator).

What is Ey, the y-component of the electric field at point P which is located at (x,y) = (0, -5 cm) as shown in the diagram?

Homework Equations


$$E\oint dA = \frac{Q_{enclosed}}{\epsilon_0}$$


The Attempt at a Solution


I'm confused on where the boundries of the shell lie, is it between the end of a and the end of b (thus making the shell have a width of b-a)?

If it IS the above, then we'd have to do analyze the problem with a Gaussian surface that ends at point P, located between the two.

We already know:
$$\rho = \frac{+Q}{\frac{4}{3}\pi r^3}$$​

$$E\oint dA = \frac{Q_{enclosed}}{\epsilon_0}$$​
$$EA = \frac{Q_{enclosed}}{\epsilon_0}$$​
$$E(4\pi r^2) = \frac{Q_{enclosed}}{\epsilon_0}$$​
$$E(4\pi r^2) = \frac{\rho \frac{4}{3}\pi r^3}{\epsilon_0}$$​
$$E = \frac{\rho r}{3\epsilon_0}$$​

However, when I attempt to use the above expression, I'm not getting a correct answer. What am I doing incorrectly, I'm definitely confused on what "r" should I be using (i.e. is "r" the .05 m given to us in the problem or is, for some reason, the sphere with radius P minus the sphere with a radius of a)?
 
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  • #2
PenDraconis said:
I'm confused on where the boundries of the shell lie, is it between the end of a and the end of b (thus making the shell have a width of b-a)?
Right.

We already know:
$$\rho = \frac{+Q}{\frac{4}{3}\pi r^3}$$​
That is an equation for a uniform sphere, not for a spherical shell.
What is the volume of the spherical shell?

You'll need another volume of a (different) spherical shell for the electric field afterwards.
 
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  • #3
Thank you!

I understood my error basically as soon as I wrote out the question despite struggling for a while.

For anyone else with a similar problem looking at this, basically there's only charge in the "shell" between b and a. So to find the charge density you'd have to do the equation for a shell with radius b and subtract from that the shell with radius a, basically netting you with:

$$\rho = \frac{Q}{\frac{4}{3}\pi (b^3-a^3)}$$​
 
  • #4
I keep on getting the density correct, but the Qenclosed incorrect whe tryign to solve this problem.I think I am going wrong with the radius or something
 
  • #5
shaniespins said:
I keep on getting the density correct, but the Qenclosed incorrect whe tryign to solve this problem.I think I am going wrong with the radius or something
Welcome to PF.

This is a 10 year old thread, but we can still try to help you with your question about the problem. You will need to show your work explicitly before we can help you, preferably posting your math using LaTeX. I will send you a Private Message (PM) with tips on how to use LaTeX (also see the "LaTeX Guide" link below the edit window).
 
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FAQ: Gauss' Law with a Insulating Shell

1. What is Gauss's Law with an Insulating Shell?

Gauss's Law with an Insulating Shell is a mathematical equation that relates the electric flux through a closed surface to the charge enclosed within that surface. It takes into account the presence of an insulating material surrounding the charge.

2. How is Gauss's Law with an Insulating Shell different from regular Gauss's Law?

Gauss's Law with an Insulating Shell takes into account the presence of an insulating material, which can change the distribution of charge and electric field inside the closed surface. Regular Gauss's Law assumes that there is no insulating material present.

3. What is an insulating shell and why is it important in Gauss's Law?

An insulating shell is a material that does not allow electric charges to move freely. In Gauss's Law, it is important because it can affect the distribution of charge and electric field inside the closed surface, and therefore, the electric flux through that surface.

4. What is the equation for Gauss's Law with an Insulating Shell?

The equation for Gauss's Law with an Insulating Shell is:
ΦE = Qenc0 + Qins0,
where ΦE is the electric flux through the closed surface, Qenc is the charge enclosed within the surface, and Qins is the charge on the insulating shell.

5. How is Gauss's Law with an Insulating Shell applied in real-world situations?

Gauss's Law with an Insulating Shell is commonly used in the study of electrostatics and can be applied to understand the behavior of electric fields in various insulating materials, such as dielectrics. It is also applied in the design and analysis of capacitors and other electronic devices.

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