Stewart James, Calculus, Section 6.4, Page 449, Problem 5
1 answer:
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a) Electric field inside the paint layer: zero
b) Electric field just outside the paint layer:
c) Electric field 8.00 cm outside the paint layer:
Explanation:
a)
We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:
where
E is the electric field
dS is the element of surface
q is the charge within the gaussian surface
is the vacuum permittivity
Here we want to find the electric field just inside the paint layer, so we take a sphere of radius as Gaussian surface, where
R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)
By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore
So we have
which means that the electric field inside the paint layer is zero.
b)
Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius
The area of the surface is
And since the electric field is perpendicular to the surface at any point, Gauss Law becomes
The charge included within the sphere in this case is the charge on the paint layer, therefore
So, the electric field is:
where the negative sign means the direction of the field is inward, since the charge is negative.
c)
Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.
Therefore, we have to take a Gaussian sphere of radius:
Gauss theorem this time becomes
And the charge included within the sphere is again the charge on the paint layer,
Therefore, the electric field is
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