Question

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?

Answer 1

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 = $q_{int}$ /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

Answer 2

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,

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