An electric field is E = 450 N/C i for x > 0 and E = -450 N/C i for x < 0. A cyl

Question

An electric field is E = 450 N/C i for x > 0 and E = -450 N/C i for x < 0. A cylinder of length 20 cm and radius 5 cm has its center at the origin and its axis along the x axis such that one end is at x = +10 cm and the other is at x = -10 cm.

(a) What is the outward flux through each end?
______Nm^2/C(+x end)
______Nm^2/C(-x end) need both answers!

(b) What is the outward flux through the curved surface of the cylinder?
______Nm^2/C

(c) What is the net outward flux through the entire cylindrical surface?
______Nm^2/C

(d) What is the net charge inside the cylinder?
_______C

I NEED Clear explanations and Answers!
Please help me

Explanation / Answer

Electric field E = 450 N/C i for x > 0
                  E ' = -450 N/C i for x < 0
Length L = 20 cm
               = 0.2 m
Radius r = 5 cm
              = 0.05 m
Area of cross section of the cylinder A = r 2
                                                           = 7.853 x 10 -3 m 2
Curved Area of the cylinder A ' = r 2 L
                                                = 1.57 x 10 -3 m 2
(a) The outward flux through +x end = EA cos 0
                                                            = 3.534 Nm 2/C
The outward flux through -x end ' = E' A cos 180
                                                       = 3.534 Nm 2/C
(b)The outward flux through the curved surface of the cylinder " = 0 Nm 2/C
Since normal to the curved surface and electric field are perpendicular to each other.
So, flux = EA ' cos 90
             = 0         Since cos 90 = 0

(c) The net outward flux through the entire cylindrical surface "' = + ' + "
                                                                                                  = 7.068 Nm 2 / C
(d)The net charge inside the cylinder q = "'
Since from gauss law net flux "' = q /
Where = Permitivity of free space
               = 8.85 x 10 -12 C 2 / Nm 2
Subsittute values we get q = 7.986 x 10 11 C

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