Spherical Shells and Gauss' Law

Just remember to include the inner shell in the partial outer shell calculation. So it would be:total inner shell, (1/4*pi*epsilon_0)*(q/r^2), + partial outer shell, [-q*(r^3 – d^3 - (b^3-a^3))]/[4*pi*epsilon*r^2*(d^3-c^3)]In summary, the problem involves a concentric arrangement of two insulating spherical shells, one with uniform charge q distributed over its volume and the other with charge -q distributed over its volume. The electric field and direction at different points within and outside of the shells are to be calculated. Using the given equations and working through the calculations, the following
  • #1
Soaring Crane
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0

Homework Statement



A small, insulating, spherical shell with inner radius a and outer radius b is concentric with a larger insulating spherical shell with inner radius c and outer radius d. The inner shell has total charge q distributed uniformly over its volume, and the outer shell has charge -q distributed uniformly over its volume.

http://i131.photobucket.com/albums/p289/SoaringCrane/yf_Figure_22_391.jpg

Calculate the magnitude of the electric field and direction of the field (outward or toward the center) for

i. a < r < b
ii. b < r < c
iii. c < r < d


Homework Equations



See below.

The Attempt at a Solution


--------------------------------------------------------------------------
General workings for shell with inner radius a and outer radius b:

Q = rho*(4/3)*pi[b^3 – a^3], where Q = total uniform charge

indefinite integral[E*dA] = q_enclosed/epsilon_0

E*4*pi*r^2 = [rho*(4/3)*pi[r^3 – a^3]]/[epsilon_0]
--------------------------------------------------------------------------

Note that then rho for inner shell = q/[(4/3)*pi*(b^3-a^3)] and outer shell rho = [-q]/[(4/3)*pi*(d^3-c^3)] in this specific problem.

a. E = [q/(4*pi*epsilon_0)]*[(r^3 – a^3)/(b^3-a^3)]*(1/r^2)

The direction will be away from the center??

b. Total charge is q, so E = (1/4*pi*epsilon_0)*(q/r^2) This will be outward the center, too?

c. This one I am really unsure of—both with the direction and electric field expression since two different charges are featured.
At first, I thought it would be [-q*(r^3 – d^3)]/[4*pi*epsilon*r^2*(c^3 – d^3)], but this alone is wrong?

Would it be a sum:

[-q*(r^3 – d^3)]/[4*pi*epsilon*r^2*(c^3 – d^3)] + [q/(4*pi*epsilon_0)]*[(r^3 – a^3)/(b^3-a^3)]*(1/r^2) + (1/4*pi*epsilon_0)*(q/r^2) ?

As for direction, would it be outward, too? A positive test charge would move away from the positively charged shell and move towards the negative shell?

Any help is appreciated. Thank you.
 
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  • #2
Can anyone please help me with part iii. and check my direction answers for the other parts?

Thank you again.
 
  • #3
Soaring Crane said:
a. E = [q/(4*pi*epsilon_0)]*[(r^3 – a^3)/(b^3-a^3)]*(1/r^2)

The direction will be away from the center??

b. Total charge is q, so E = (1/4*pi*epsilon_0)*(q/r^2) This will be outward the center, too?
These look OK to me.

c. This one I am really unsure of—both with the direction and electric field expression since two different charges are featured.
At first, I thought it would be [-q*(r^3 – d^3)]/[4*pi*epsilon*r^2*(c^3 – d^3)], but this alone is wrong?
Here it seems like you just considered the outer shell of charge but forgot to include the inner shell.

Would it be a sum:

[-q*(r^3 – d^3)]/[4*pi*epsilon*r^2*(c^3 – d^3)] + [q/(4*pi*epsilon_0)]*[(r^3 – a^3)/(b^3-a^3)]*(1/r^2) + (1/4*pi*epsilon_0)*(q/r^2) ?
It would be the sum of the field from the partial outer shell plus the field from the total inner shell--you've computed them both, just add them up.

As for direction, would it be outward, too? A positive test charge would move away from the positively charged shell and move towards the negative shell?
The field at point r just depends on the total charge within the spherical surface at r. In this problem, that total charge is always positive for all points up to r = d, thus it points outward at all points.
 
  • #4
The sum will be:


total inner shell, (1/4*pi*epsilon_0)*(q/r^2), + partial outer shell,
[-q*(r^3 – d^3)]/[4*pi*epsilon*r^2*(c^3 – d^3)] ?

Or did I use/mismatch the wrong expressions?
 
  • #5
That looks good to me.
 

1. What is a spherical shell in physics?

A spherical shell in physics is a three-dimensional surface that encloses a certain amount of space. It is a common object used in physics to represent various concepts, such as electric fields, gravitational fields, and pressure.

2. How is Gauss' Law applied to spherical shells?

Gauss' Law is a fundamental law in physics that relates the electric field at a point to the charge enclosed by a surface. In the case of a spherical shell, Gauss' Law states that the electric field at any point outside the shell is equal to the total charge of the shell divided by the square of the distance from the center of the shell.

3. What is the significance of spherical shells in electrostatics?

Spherical shells are important in electrostatics because they allow us to simplify complex charge distributions into a single point charge at the center of the shell. This greatly simplifies the calculations of electric fields and forces in many practical situations.

4. Can Gauss' Law be applied to non-spherical shapes?

Yes, Gauss' Law can be applied to non-spherical shapes as long as the charge distribution is symmetric. In these cases, the electric field at any point can still be calculated using the enclosed charge and the distance from the center of the shape.

5. How does the electric field inside a charged spherical shell differ from the electric field outside the shell?

The electric field inside a charged spherical shell is always zero. This is because the charge on the shell is distributed evenly on the surface, so there is no net electric field inside. Outside the shell, the electric field follows the inverse square law and decreases as the distance from the center of the shell increases.

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