Answer:
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
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