The electric field in a particular space is = (x + 3.6) N/C with x in meters. Consider a cylindrical Gaussian surface of radius

Answer:

a) Ф = 0.016 N / C m, b) q_{int} = 0.14 10⁻¹² C

Explanation:

a) For this scenario, we rely on Gauss's law

          Ф = E.ds = /ε₀

As the field points in the x direction, there is no flux 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

Now, let's compute

          Ф = (3.7 -3.6) 0.16

          Ф = 0.016 N / C m

b) Using Gauss's law, we have

             q_{int} = Ф ε₀

Where the flow is present on both sides, at the face corresponding to x = 0, the flow is zero

             q_{int} = 0.016 8.85 10⁻¹²

             q_{int} = 0.14 10⁻¹² C