What is the electric field on a thin nonconducting shell

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Homework Help Overview

The discussion revolves around a problem involving a nonconducting thin spherical shell with a uniform surface charge density. Participants are exploring the electric field at various distances from the shell and calculating the total charge on the shell.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of total charge based on surface charge density and the implications of Gauss's law for determining the electric field at different radial distances from the shell.

Discussion Status

Some participants have confirmed their calculations and are seeking validation of their reasoning, particularly regarding the electric field inside and outside the shell. There is an ongoing exploration of assumptions related to charge distribution and electric field behavior.

Contextual Notes

Participants are considering the significance of significant figures in their final answers and questioning the correctness of their logic in applying Gauss's law to the problem.

henrco
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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 Gauss'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 Gauss'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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