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The angular speed is defined as:
<h2> ω=where
A) The electric field inside the paint layer is zero
B) The electric field just outside the paint layer is (radially inward)
C) The electric field at 6.00 cm from the surface is (radially inward)
Explanation:
A)
We can solve the problem by applying Gauss Law, which states that the electric flux through a Gaussian surface must be equal to the charge contained in the surface divided by the vacuum permittivity:
where
E is the magnitude of the electric field
dS is the element of the surface
q is the charge contained within the surface
is the vacuum permittivity
By taking a sphere centered in the origin,
where is the surface of the Gaussian sphere of radius r.
In this problem, we want to find the electric field just inside the paint layer, so we take a value of r smaller than
(radius of the plastic sphere is half of the diameter)
Since the charge is all distributed over the plastic sphere, the charge contained within the Gaussian sphere is zero:
And therefore,
So, the electric field inside the plastic sphere is zero.
B)
Here we apply again Gauss Law:
In this case, we want to calculate the electric field just outside the paint layer: this means that we take r as the radius of the plastic sphere, so
The charge contained within the Gaussian sphere is therefore
Therefore, the electric field is
And the negative sign indicates that the direction of the field is radially inward (because the charge that generates the field is negative). However, the text of the question says "Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward", so the answer to this part is
C)
For this part again, we apply Gauss Law:
In this case, we want to calculate the field at a point 6.00 cm outside the surface of the paint layer; this means that the radius of the Gaussian sphere must be
r = 9 cm + 6 cm = 15 cm = 0.15 m
While the charge contained within the sphere is again
Therefore, the electric field in this case is
And again, this is radially inward, so according to the sign convention asked in the problem,
Learn more about electric fields:
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