What is the electric flux through a surface with a point charge at its center?

In summary, 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 sock comes out of the dryer with 10^12 excess electron. What's the electric flux through a surface surrounding the socks?
  • #1
katta002
5
0

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.
 
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  • #2
Welcome to PF!

Hi katta002! Welcome to PF! :smile:
katta002 said:
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


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
katta002 said:
… 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)?​
 
  • #5
pi*r^2
 
  • #6
katta002 said:
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:

1. What is electric flux of a surface?

The electric flux of a surface is a measure of the total electric field passing through a given area. It is represented by the symbol Φ and is measured in units of volts per meter squared (V/m²).

2. How is electric flux calculated?

Electric flux is calculated by multiplying the electric field strength by the perpendicular area of the surface. The formula for electric flux is Φ = E x A x cos(θ), where E is the electric field strength, A is the area of the surface, and θ is the angle between the electric field and the surface.

3. What is the significance of electric flux?

Electric flux plays an important role in understanding the flow of electric fields. It helps in determining the strength and direction of electric fields passing through a given surface, which is useful in many practical applications such as designing electrical circuits and understanding the behavior of charged particles.

4. How does a change in surface area affect electric flux?

A larger surface area results in a greater amount of electric flux, as there is more area for the electric field to pass through. Similarly, a smaller surface area will result in a lower electric flux. However, the direction of the electric field and the angle between the field and the surface also play a role in determining the electric flux.

5. Can electric flux be negative?

Yes, electric flux can be negative. This occurs when the electric field and the surface have opposite directions, resulting in a negative value for the electric flux. This indicates that the electric field is entering the surface rather than exiting it.

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