Question

An electric field is E = sin(x)·(420 N/C)i, where sin(x) = −1 if x < 0, 0 if x = 0, and +1 if x > 0. A cylinder of length L = 22 cm and radius R = 4.5 cm has its center at the origin and its axis along the x-axis such that one end is at x= +11 cm and the other is at x = -11 cm.
(a) What is the flux through each end?
(b) What is the flux through the curved surface of the cylinder?
(c) What is the net outward flux through the entire cylindrical surface?
(d) What is the net charge inside the cylinder?

Answer 1

Answer:

a) +2.67192 Nm^2 / C , +2.67192 Nm^2 / C

b) 0

c) 5.34384 Nm^2/C

d) 4.73*10^-11 C

Explanation:

Given:

E =  sign (x) * 420 N / C i

E = + 420 N/C i      x >0

E = 0                        x=0

E = -420 N/C i        x<0

L = 22 m

R = 0.045 m

Find:

a) flux through each end.

We have two ends with respective Electric field strength given by the function above. We will apply Gauss Law at the two surfaces

       Flux @ face x = + 11 m

       Θ = $\int\limits^S {E} \, .dA$ = + 420 N/C i . (pi*R^2) i = 420 * (pi * 0.045^2)

       Θ = +2.67192 Nm^2 / C

       Flux @ face x = - 11 m

       Θ = $\int\limits^S {E} \, .dA$ = - 420 N/C i .  - (pi*R^2) i = 420 * (pi * 0.045^2)

       Θ =  +2.67192 Nm^2 / C

(b) Flux through the curved surface of the cylinder

We have two parts of curved surfaces with respective Electric field strength given by the function above. We will apply Gauss Law at the two surfaces.

       Flux @ curved face 0 <  x < + 11 m and - 11 < x < 0

       Θ = $\int\limits^S {E} \, .dA$ = + 420 N/C i . (2*pi*R*11) r = 0

       Θ = $\int\limits^S {E} \, .dA$ = - 420 N/C i . (2*pi*R*11) r = 0

       The normal vector dA = 2*pi*R*dx r points radially out-wards. While E         = +420 i points parallel to the curved surface. The Dot product of two   orthogonal vectors is 0. Hence, Flux through curved surface is also zero.

(c) net outward flux through the entire cylindrical surface

The net flux of the entire cylindrical surface is the sum of all flux through part of the surfaces i.e two faces and curved surface. Hence,

         Θ_net = Θ_+face + Θ_-face + Θ_curved

         Θ_net = + 2.67192 + 2.67192 + 0

         Θ_net = 5.34385 Nm^2 / C

(d) Net charge inside a closed surface:

According to Gauss Law the net flux of a closed surface is equal to the enclosed charge divided the permittivity of space constant.

                             Θ =  $\int\limits^S {E} \, .ds = \frac{Q_e}{e_o}$

                             Q_e = Θ_net * e_o

                             Q_e = (5.34385 Nm^2/C) * 8.85*10^-12

                             Q_e = 4.73 * 10^-11 C