# Homework Help: Electric flux calculation in case of a cylinder

1. Sep 13, 2015

### gracy

1. The problem statement, all variables and given/known data

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 vector E=200 I^ (N/C) for x>0 and vector E= -200 (N/C) for x<0 .A right circular cylinder of length 20 cm and radius 5 cm has it's center at the origin and it's axis along the x-axis so that one face is at x=+10 cm and the other is at x=-10 cm
a)what is the flux through the side of the cylinder?
b)what is the net outward flux through the cylinder?

2. Relevant equations

3. The attempt at a solution
a)in my book it has been given that Vector E is parallel to sides of cylinder so flux through sides =0
My understanding regarding this explanation of Vector E to be parallel to sides of cylinder is to point out that that electric fields do not even pass through sides of cylinder because the angle between vector E and area vector of sides of cylinder is 90 degrees so no flux.Am I correct?
b)net outward flux=flux through faces (as flux through sides is zero)
=vector E.vector A +vector 'E.vector A
=0
But solution of this is given to be
=2EA
=3.14 Nm^2/C
I don't understand why?why magnitude of 'E is taken as 200 N/C but not -200 N/C
I thought negative sign is part of magnitude because it does not symbolize/indicate direction as direction is specified by vector i.

2. Sep 13, 2015

### #-Riley-#

Well it sounds like you've got an infinite plane of charge. You're right in saying that the flux throug h the sides of the cylinder is zero because the field is everywhere tangent to the sides. The end faces are perpendicular to the field and the field is uniform so $$\oint \vec E \bullet d \vec A$$ just becomes EA. But the cylinder has two ends, and the vector A is in the same direction as the field in both cases (away from the y axis) so the flux through each end is EA. You have two lots of EA. i.e Total flux = 2EA

The important you're missing is that the question is asking for the flux 'out' of the cylinder. The field exits the cylinder at both ends meaning there is charge enclosed within the surface and so there is a net flux out of the cylinder. If there was no charge enclosed, then the flux in one end would equal the flux out the other and there would be zero net flux.

3. Sep 13, 2015

### gracy

4. Sep 13, 2015

### gracy

5. Sep 13, 2015

### #-Riley-#

The negative sign is simply giving you the direction of the field so you can compare that with the direction of the vector A. The magnitude is simply the absolute value of E. The strength of the field, not concerned with direction.

Think of the flux through the end of the cylinder as being like water flowing out the end of a pipe. If you there is water flowing out both opposite ends of the pipe, would you say the total water flowing out of the pipe was zero? Or would you say it is double the water flowing out one end?

Hopefully you took the second option then, and hopefully you noticed that the fact that the water is coming out in two opposite directions, is irrelevant. they dont cancel each other out

6. Sep 13, 2015

### gracy

Then what i^ is for?

7. Sep 13, 2015

### #-Riley-#

Well it just goes along with the negative sign to complete the direction. So that you know the field is in completely opposite directions but equal in magnitude on each side of the y axis. Which is the reason you can model it as an infinite plane of charge.

8. Sep 13, 2015

### #-Riley-#

The negative sign and the unit vector i are just notation things. Literally just a mathematical way of saying exactly what your first line was

9. Sep 13, 2015

### gracy

so i^ denotes it's about x axis and negative sign tells more precisely that's it is regarding negative x axis,right?

10. Sep 13, 2015

### #-Riley-#

Yep! Exactly :)