Question

Example 1.12 An electric field is uniform, and in the positive x
direction for positive x, and uniform with the same magnitude but in
the negative x direction for negative x. It is given that E = 200 i N/C
for x > 0 and E = -200 i N/C for x < 0. A right circular cylinder of
length 20 cm and radius 5 cm has its centre at the origin and its axis
along the x-axis so that one face is at x = +10 cm and the other is at
x = -10 cm (Fig. 1.28). (a) What is the net outward flux through each
flat face? (b) What is the flux through the side of the cylinder?
(c) What is the net outward flux through the cylinder? (d) What is the
net charge inside the cylinder?​

Answer 1

Explanation:

(a) net outward flux through each  flat face is given by

$\vec{E}.\vec{A}\\\\here \\\\\theta = 0^\circ\\\\EA\cos\theta\\\\(200)(\pi(0.05)^2)\\\\=\frac{\pi}{2}\frac{NM^2}{c^2}\\\\1.5707\frac{NM^2}{c^2}$

(b) flux through the side of the cylinder is 0 since E⊥A

no electric field is perpendicular to the curved surface of the cylinder

(c) net outward flux through the cylinder

$=\pi\frac{NM^2}{c^2}\\\\3.14\frac{NM^2}{c^2}$

(d) net charge q inside the cylinder  = net outward flux of closed surface =

$\frac{q}{\epsilon _0}\\\\q = \pi \times \epsilon _0\\\\\pi \times 8.854\times 10^-^1^2\\\\= 27.82\times 10^-^1^2 C$

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Answer 2

Net flux

Explanation:

(a) Net ϕ through flat surface = E.A (where θ is zero)

= EACosθ

= 200(π0.05²)

= πNm²/2C²

= 1.5707Nm²/C²

(b) Net ϕ through curved surface = E.A (where θ is 90)

 = 0

(c) Net outward flux = 2 * ϕ flat surface

= πNm²/C²

= 3.14Nm²/C²