# What is the electric field on a thin nonconducting shell

1. Feb 10, 2016

### henrco

1. The problem statement, all variables and given/known data
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.

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

and Guass's law

3. The attempt at a solution

Part i) What is the total charge on the shell?

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?

2. Feb 10, 2016

### Dr. Courtney

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.

3. Feb 11, 2016

### henrco

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.

4. Feb 11, 2016

### henrco

Solved this in the end. My calculations were all correct.