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Magnitude of a Net Charge within a closed surface

  • Thread starter fisixC
  • Start date
  • #1
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Homework Statement



A closed surface with dimensions a = b =
0.254 m and c = 0.4064 m is located as in
the figure. The electric field throughout the
region is nonuniform and given by ~E = (α +
β x^2)ˆı where x is in meters, α = 4 N/C, and
β = 6 N/(C m2).

Picture of object attached.

What is the magnitude of the net charge
enclosed by the surface?
Answer in units of C.

Homework Equations



Gauss's Law: Flux = integral(E dA) = Q/permittivity constant

permittivity constant(epsilon naught) = 8.85E-12

The Attempt at a Solution



I know that the object doesn't not have any flux through any of the sides except the left and right sides(those parallel to the electric field). So I thought to find net flux and multiply that by the permittivity constant and find the net charge enclosed.

Flux1 = (α +β (a+c)^2)*(a*b) = Electric field * area of the side(right)
Flux2 = (α +β (a)^2) * (a*b) = Electric field * area of the other side(left)

Then:
Flux1 + Flux2 = 0.709925351759
Therefore I multiply by epsilon naught:
net flux * epsilon naught = 6.28283939307E-12

However this is wrong, I believe I am messing up because of the weird way they are giving the electric field as, or possibly my entire calculations or process?
 

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Answers and Replies

  • #2
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Anyone, please help.
 
  • #3
23
1
Are you sure you should be adding the fluxes? Knowing that flux is the integral of E dotted into dA, what can you say about the flux for either side?
 
  • #4
54
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I figured it out I needed to subtract the positive flux from the negative flux. Right side - left side.
 
  • #5
23
1
Right. Do you know why?
 
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Likes idkjayesh
  • #6
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Yes because the area vector is going the same direction as the electric field vector on the right but on the left the area and electric field vectors are going opposite directions.
 
  • #7
23
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Excellent. Keep at the physics! It only gets more fun from where you are, trust me.
 
  • #8
gneill
Mentor
20,793
2,773
For one of the sides the normal to that side is antiparallel to the field lines (opposite direction), while for the other side the normal is parallel to it. Thus one flux should have a negative value.

EDIT: Well, I see I'm a bit late to the party. Glad to see you worked it out!
 

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