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Electic Flux through a cube Why am I getting a value?

  • Thread starter Tautomer22
  • Start date
  • #1

Homework Statement



A cubical Gaussian surface is placed in a uniform electric field as shown in the figure to the right. The length of each edge of the cube is 1.0 m. The uniform electric field has a magnitude of 5.0 * 10^8 N/C and passes through the left and right sides of the cube perpendicular to the surface. What is the total electric flux that passes through the cubical Gaussian surface?

http://www.ahsd.org/science/stroyan/APPhysics/CH15/estat/f18051.jpg [Broken]


A. 5.0 ´ 108 N•m2/C
B. 3.0 ´ 109 N•m2/C
C. 2.5 ´ 106 N•m2/C
D. 1.5 ´ 107 N•m2/C
E. zero N•m2/C

Homework Equations



I realize that electric flux is ƩE*Cos∅*ΔA = q/permitivity of free space

My physics books states that the angle in question is the difference in degrees between the NORMAL of the surface and the electric field.

In the case of a cube here, I would assume the two faces to which the electric field is perpendicular would thus have cos (0) or a value 1 which would reduce the electric flux to: ƩE*ΔA

So, there should be two sides that would give me some sort of value.

The Attempt at a Solution



I simply followed the equation through for one side and got an electric flu equal to the magnitude of the elecrtic field with a m^2 valued tacked on to it. This is answer A here.

However,the correct answer is apparently E, or an electric flux of 0. I do not understand this, because I don't see how the electric field could be 90 degrees to the normal of the cube surface for every face shown, since at least 2 of the faces need to have a value for the electric flux.

Could someone please help me? I have this gnawing feeling that I'm missing something extremely simple.
 
Last edited by a moderator:

Answers and Replies

  • #2
gneill
Mentor
20,793
2,773

Homework Statement



A cubical Gaussian surface is placed in a uniform electric field as shown in the figure to the right. The length of each edge of the cube is 1.0 m. The uniform electric field has a magnitude of 5.0 * 10^8 N/C and passes through the left and right sides of the cube perpendicular to the surface. What is the total electric flux that passes through the cubical Gaussian surface?

http://www.ahsd.org/science/stroyan/APPhysics/CH15/estat/f18051.jpg [Broken]


A. 5.0 ´ 108 N•m2/C
B. 3.0 ´ 109 N•m2/C
C. 2.5 ´ 106 N•m2/C
D. 1.5 ´ 107 N•m2/C
E. zero N•m2/C

Homework Equations



I realize that electric flux is ƩE*Cos∅*ΔA = q/permitivity of free space

My physics books states that the angle in question is the difference in degrees between the NORMAL of the surface and the electric field.

In the case of a cube here, I would assume the two faces to which the electric field is perpendicular would thus have cos (0) or a value 1 which would reduce the electric flux to: ƩE*ΔA

So, there should be two sides that would give me some sort of value.

The Attempt at a Solution



I simply followed the equation through for one side and got an electric flu equal to the magnitude of the elecrtic field with a m^2 valued tacked on to it. This is answer A here.

However,the correct answer is apparently E, or an electric flux of 0. I do not understand this, because I don't see how the electric field could be 90 degrees to the normal of the cube surface for every face shown, since at least 2 of the faces need to have a value for the electric flux.

Could someone please help me? I have this gnawing feeling that I'm missing something extremely simple.
Hi Tautomer22. Welcome to Physics Forums.

Keep in mind that normals are vector quantities and have direction. While one of those outward-facing normals will point along the +x-axis, the other will point in the opposite direction...
 
Last edited by a moderator:
  • #3
Hi Tautomer22. Welcome to Physics Forums.

Keep in mind that normals are vector quantities and have direction..
I was unclear on this point, thank you so much gneil! That solves the whole issue for me. I knew I was missing a basic idea here, and I wasn't deducing it from solutions to similar problems (alas).
 

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