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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 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?