Electric flux of surface

  • Thread starter katta002
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  • #1
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Homework Statement


A flate surface with area 0.14m^2 lies in x-y plane in a uniform electric field given by E=5.1i+2.1j+3.5k kN/c

find the flux through this surface


Homework Equations



I know that flux= EA which E and A are given so I multiply The area with the victor E but I got a wrong answer

The Attempt at a Solution


0.14m^2(5.1i+2.1j+3.5k kN/c)=7.14e-1i+2.94e-1j+4.9e-1z

but the correct answer is 490Nm^2/C,

how come can someone please help me. thanks.




Homework Statement


3)A 10-{ nC} point charge is located at the center of a thin spherical shell of radius 8.0 { cm} carrying -20 {{ nC}} distributed uniformly over its surface.

A.What is the magnitude of the electric field 2.0 {cm} from the point charge and the direction of the field?

B. What is the magnitude of the electric field 15 { cm} from the point charge?

Homework Equations


E outside spherical =Q/4pi[tex]k\ =\ 1.3806503(24)\ \times\ 10^{-23}\ J\ K^{-1}[/tex]r^2

and E inside= Qr/4pi[tex]k\ =\ 1.3806503(24)\ \times\ 10^{-23}\ J\ K^{-1}[/tex]r^3


The Attempt at a Solution

R^3
A)and since 8.0cm>2.0cm it's outside so I used the outside equation
(10e19)(2e-2)/(4pi(9e-2)(8e2^3)=3.45e3

but the correct answer is 230kN/C

b) since 15cm>8.0cm the E is inside therefore
(.08)/4pi(9e-12)(.15)=4.19.e3

but my answer is wrong

Homework Statement


2)A sock comes out of the dryer with 10^12 excess electron.
What's the electric flux through a surface surrounding the socks?

Homework Equations


Flux= EA
but since the area was not given
flux= Qenclosed?[tex]k\ =\ 1.3806503(24)\ \times\ 10^{-23}\ J\ K^{-1}[/tex]

The Attempt at a Solution


10^12.9e-12=1.e23

but correct answer is 1.18e4Nm^2/C

please help me because I have a test on flux tomorrow and without understanding this might cause me. thanks.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

Hi katta002! Welcome to PF! :smile:
A flate surface with area 0.14m^2 lies in x-y plane in a uniform electric field given by E=5.1i+2.1j+3.5k kN/c

find the flux through this surface

I know that flux= EA which E and A are given so I multiply The area with the victor E but I got a wrong answer

0.14m^2(5.1i+2.1j+3.5k kN/c)=7.14e-1i+2.94e-1j+4.9e-1z

but the correct answer is 490Nm^2/C
You need, not the flux through the area, but the component of flux in the direction normal (perpendicular) to the area.

So the normal to the area is … ?, and the component of flux in that direction is … ? :smile:

(and how are you doing on the others?)​
 
  • #3
5
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Thank you for welcoming me PF and boy am i glad to find this website
So I multiply the area by the i direction, j direction and K direction and find the sum of all direction but still didn't get the answer
i=.714
j=.294
k=.49
the sum is 1.498, I still don't get it.

and the rest that I posted I'm still facing difficulties so could u break it down for me, thanks and God bless you
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
… So I multiply the area by the i direction, j direction and K direction and find the sum of all direction
that's not the component in the normal direction …

it's not even a dot-product of anything.

What is the normal direction (of the area)?​
 
  • #6
tiny-tim
Science Advisor
Homework Helper
25,832
251
pi*r^2
uhh? :confused:

The plate lies in the x-y plane …

so the normal direction is the z-direction (k) …

so the question is asking what is the component of flux in the k-direction?

(btw, did you find out how to post in the academic guidance forum?)
 
Last edited:

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