Magnitude of a Net Charge within a closed surface

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Homework Help Overview

The problem involves calculating the magnitude of the net charge enclosed by a closed surface in the presence of a nonuniform electric field. The electric field is defined as ~E = (α + β x^2)ˆı, with specific values for α and β, and the dimensions of the surface are provided.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of electric flux through the surface and question the method of adding fluxes from different sides of the surface. There is a focus on understanding the direction of the electric field relative to the area vectors of the surface.

Discussion Status

The discussion has progressed with some participants clarifying the need to consider the direction of the electric field when calculating flux. There is acknowledgment of a mistake in the original approach, leading to a more accurate understanding of how to handle the flux from opposing sides of the surface.

Contextual Notes

Participants are working under the constraints of Gauss's Law and are exploring the implications of the electric field's nonuniformity on the calculation of net charge. There is an emphasis on understanding the physical principles rather than simply applying formulas.

fisixC
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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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Anyone, please help.
 
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?
 
I figured it out I needed to subtract the positive flux from the negative flux. Right side - left side.
 
Right. Do you know why?
 
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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.
 
Excellent. Keep at the physics! It only gets more fun from where you are, trust me.
 
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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