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28.02.2020 •
Physics
The electric field in a particular space is = (x + 3.6) N/C with x in meters. Consider a cylindrical Gaussian surface of radius 16 cm that is coaxial with the x axis. One end of the cylinder is at x = 0. (a) What is the magnitude of the electric flux through the other end of the cylinder at x = 3.7 m? (b) What net charge is enclosed within the cylinder?
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Ответ:
The magnitude of the electric flux is 0.016 N / C m and net charge enclosed within the cylinder is 0.14 10⁻¹² C.
a) From the Gauss's law
Ф = E .ds
Ф = Qnet/ε₀
The electric flux,
Ф = E A
Where,
A - area = π r
Thus,
Ф = E π r
Ф = (x- 3.6) r
Ф = (3.7 -3.6) 0.16
Ф = 0.016 N / C m
b) From Gauss's law
Qnet = Ф ε₀
Qnet = 0.016 8.85 10⁻¹²
Qnet = 0.14 10⁻¹² C
To know more about Gauss's law,
link
Ответ:
a) Ф = 0.016 N / C m , b) q_{int} = 0.14 10⁻¹² C
Explanation:
a) For this problem we use Gauss's law
Ф = E .ds =
/ε₀
The camp is in the x direction so it has no flow through the cylinder walls.
Ф = E A
The area of a circle is
A = π r
Ф = E π r
Ф = (x- 3.6) r
Let's calculate
Ф = (3.7 -3.6) 0.16
Ф = 0.016 N / C m
b) we clear from Gauss's law
q_{int} = Ф ε₀
Where the flow is on both sides, on the face at x = 0 the flow is zero
q_{int} = 0.016 8.85 10⁻¹²
q_{int} = 0.14 10⁻¹² C
Ответ: