Electric field of spherical shell

In summary, the conversation discusses the calculation of the electric field at different distances from a spherical shell with uniform charge density. The solution uses the volume of the shell to calculate the total charge, while the question mentions using the area of the shell. The concept of a spherical shell and its volume between two spherical surfaces is explained. The final formula for the electric field is also provided.
  • #1
starstruck_
185
8

Homework Statement


Consider a spherical shell with uniform charge density ρ.
The shell is drawn as a donut with inner (R1) and outer (R2) radii.
Let r measure the distance from the center of the spherical shell, what is the electric field at r>R2, R1<r<R2, and r<R1.

I am working on the r > R2 part right now.
I'll work on R1<r<R2 later.

The question says spherical shell so I would have assumed that we would multiply ρ by the area of the shell instead of the volume, however my professor's solution uses volume for the total charge. I solved the problem, I just don't understand why we would use volume and not the shell area. Although, it DOES give ρ which is the density per unit of volume.
It just doesn't make sense to me. What am I missing? Am I misunderstanding something?

Homework Equations


E = Qenclosed/ (ε0A)

The Attempt at a Solution


For r > R2 [/B]
Qenclosed = ρ*4/3π(R2^3-R1^3)
A = 4πr^2
R2 is the outer radius of the shell and r is the distance from the center.

E = ρ(R2^3-R1^3)/(3ε0*r^2)
I already know that r<R1 =0 from the center of the shell to R1, there is no charge (it is a hole).
 
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  • #2
The charge is spread throughout a volume. Imagine two spheres with the same center. One sphere has radius R1 and the other has radius R2. The charge is spread throughout the volume between the surfaces of the two spheres. This region between the two spherical surfaces constitutes the "spherical shell".

See figure (a) here
https://www.chegg.com/homework-help/questions-and-answers/figure-b-nonconducting-spherical-shell-inner-radius-206-cm-outer-radius-b-256-cm-volume-ch-q3674386
 
  • #3
TSny said:
The charge is spread throughout a volume. Imagine two spheres with the same center. One sphere has radius R1 and the other has radius R2. The charge is spread throughout the volume between the surfaces of the two spheres. This region between the two spherical surfaces constitutes the "spherical shell".

See figure (a) here
https://www.chegg.com/homework-help/questions-and-answers/figure-b-nonconducting-spherical-shell-inner-radius-206-cm-outer-radius-b-256-cm-volume-ch-q3674386
AHH right! That makes sense! Thank you (totally forgot about that from calculus)
 

1. What is the formula for the electric field of a spherical shell?

The formula for the electric field of a spherical shell is E = kQ/r², where k is the Coulomb's constant, Q is the total charge of the shell, and r is the distance from the center of the shell to the point where the electric field is being measured.

2. How does the electric field of a spherical shell vary with distance?

The electric field of a spherical shell varies inversely with the square of the distance from the center of the shell. This means that as the distance increases, the electric field decreases.

3. Is the electric field inside a spherical shell zero?

Yes, the electric field inside a spherical shell is always zero. This is because the charge on the shell is distributed evenly on the outer surface, and there is no charge inside the shell to create an electric field.

4. How does the electric field of a spherical shell compare to that of a point charge?

The electric field of a spherical shell is similar to that of a point charge when you are outside the shell. However, inside the shell, the electric field of a point charge is non-zero, while the electric field of a spherical shell is always zero.

5. Can the electric field of a spherical shell be negative?

Yes, the electric field of a spherical shell can be negative. This occurs when the charge on the shell is negative, and the electric field points towards the center of the shell. However, the magnitude of the electric field will still follow the inverse square law and decrease as the distance from the shell increases.

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