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Calculating Electric Flux

  • #1

Homework Statement


A 2cm x 3cm rectangle lies in the xy plane. What is the electric flux through the rectangle if
Electric field= (100i +50k) N/C

Homework Equations


Φe=E⋅Acosθ (Electric Flux Equation)

The Attempt at a Solution



My question is to find the magnitude of the electric field we say Emag=√(100i)2+(50k)2= 111.8N/C
Area=6x10-4m2
To find direction θ=tan-(50/100)=26.5°
We know that The angle in formula for electric flux is the angle subtended from a line normal to the surface and the electric field line. Therefore 90°-26.5°=63.4°
so:
Φe=E⋅Acosθ=(111.8N/C)(6x10-4m2)cos(63.4)=3.0x10-2

The textbook is disagreeing with this answer... Not sure what I am doing wrong.
 

Answers and Replies

  • #2
kuruman
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The calculation is more straightforward if you consider that only the component that is perpendicular to the surface contributes to the flux. Nevertheless, your answer seems OK to me.
 
  • #3
collinsmark
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Homework Statement


A 2cm x 3cm rectangle lies in the xy plane. What is the electric flux through the rectangle if
Electric field= (100i +50k) N/C

Homework Equations


Φe=E⋅Acosθ (Electric Flux Equation)

The Attempt at a Solution



My question is to find the magnitude of the electric field we say Emag=√(100i)2+(50k)2= 111.8N/C
Area=6x10-4m2
To find direction θ=tan-(50/100)=26.5°
We know that The angle in formula for electric flux is the angle subtended from a line normal to the surface and the electric field line. Therefore 90°-26.5°=63.4°
so:
Φe=E⋅Acosθ=(111.8N/C)(6x10-4m2)cos(63.4)=3.0x10-2

The textbook is disagreeing with this answer... Not sure what I am doing wrong.
What answer does the textbook give? By the way, for what it's worth, I agree with your numerical answer. A bit more below though about your method.

While I agree with the answer that you obtained, you made this way too complicated.

It's true that

[itex] \vec E \cdot \vec A = EA \cos \theta. [/itex]

But there is another way to express this relationship:

[itex] \vec E \cdot \vec A = E_x A_x + E_y A_y + E_z A_z, [/itex]

which is far more useful here.

You need to know both of these relationships. Sometimes one will be easier, and some other times the other will. You need to know both.
 
  • #4
Ah
What answer does the textbook give? By the way, for what it's worth, I agree with your numerical answer. A bit more below though about your method.

While I agree with the answer that you obtained, you made this way too complicated.

It's true that

[itex] \vec E \cdot \vec A = EA \cos \theta. [/itex]

But there is another way to express this relationship:

[itex] \vec E \cdot \vec A = E_x A_x + E_y A_y + E_z A_z, [/itex]

which is far more useful here.

You need to know both of these relationships. Sometimes one will be easier, and some other times the other will. You need to know both.
Ahhh okay, makes it much easier. The book says 3.2x10^-5 Nm^2/C
 
  • #5
The calculation is more straightforward if you consider that only the component that is perpendicular to the surface contributes to the flux. Nevertheless, your answer seems OK to me.
Thank you!
 
  • #6
collinsmark
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The book says 3.2x10^-5 Nm^2/C
Yeah, I suspect there is a mistake in your textbook, somewhere or another.
 

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