What is the electric field on a thin nonconducting shell

AI Thread Summary
A nonconducting thin spherical shell with a radius of 6.80 cm and a surface charge density of 9.28 nC/m² has a total charge of 5.39 x 10^-10 C. The electric field inside the shell, at radii of 2.43 cm and 6.61 cm, is zero. At a radius of 7.17 cm, the electric field is calculated to be 942 N/C, while at 11.3 cm, it is 379 N/C, demonstrating a decrease in electric field strength with distance from the shell. The calculations for the total charge and electric fields are confirmed to be correct, and the reduction of the final answer to three significant figures is deemed appropriate. The logic applied in the calculations aligns with Gauss's law principles.
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Homework Statement


A nonconducting thin spherical shell of radius 6.80 cm has a uniform surface charge density of 9.28 nC/m2.

i) What is the total charge on the shell?
ii) Find the electric field at r = 2.43 cm.
iii) Find the electric field at r = 6.61 cm.
iv) Find the electric field at r = 7.17 cm.
v) Find the electric field at r = 11.3 cm.

Homework Equations


p = Q/V (where p = charge density, Q = total charge and V = volume.

and Guass's law

The Attempt at a Solution



Part i) What is the total charge on the shell? [/B]
To determine the total charge, we have surface charge density of 9.28 nC/m2 and radius of .068m
Convert 9.28nC to C, 9.28x10^-9 and convert radius of 6.8cm to m, .068 m

We will use the equation p = q/V. Rearrange: Q = QV
The surface area of the spherical shell is 4 pi r^2

Q = pV
Q = (9.28 x 10^-9) (4 pi (.068)^2)
Q = 5.39 x 10^-10 C

Part ii) Find the electric field at r = 2.43 cm.
Inside the this spherical shell, the electric field is zero everywhere.

Part iii) Find the electric field at r = 6.61 cm.
Inside the this spherical shell, the electric field is zero everywhere.

iv) Find the electric field at r = 7.17 cm.
The flux through the Gaussian surface again is:

E dA = E (4 pi r ^2)

Applying Guass's law, we can derive this formula.

E = Q / (4 pi Eo r^2) (Where Eo is the permittivity of free space)
E = (5.39 x 10^-10)/ (4 pi Eo (.0717)^2)
E = 942 N/C

iv) Find the electric field at r = 11.3 cm.

E = Q / (4 pi Eo r^2) (Where Eo is the permittivity of free space)
E = (5.39 x 10^-10)/ (4 pi Eo (.113)^2)
E = 379 N/C

As expected a weaker electric field as you go further from the surface of the sphere.

Question : The actual answer is 379.387 etc. However I am reducing the answer to 3 significant figures. Is that correct?
 
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You can double check by assuming all the charge is located at a point at the center and computing the electric field of a point charge at the locations of interest.
 
I'm now sure part i) is correct where the total charge on the thin sphere is Q = 5.39 x 10^-10 C.

But could I please get guidance on if my logic is correct with part ii) with my earlier post.
 
Solved this in the end. My calculations were all correct.
 
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