Magnitude of a Net Charge within a closed surface

In summary, Gauss's Law states that flux is the integral of electric field across a surface. The attempt at a solution focused on finding the net charge enclosed by a surface with nonuniform electric field. The solution was found by subtracting positive and negative fluxes.
  • #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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  • #2
Anyone, please help.
 
  • #3
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
I figured it out I needed to subtract the positive flux from the negative flux. Right side - left side.
 
  • #5
Right. Do you know why?
 
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  • #6
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
Excellent. Keep at the physics! It only gets more fun from where you are, trust me.
 
  • #8
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!
 

Related to Magnitude of a Net Charge within a closed surface

1. What is the definition of magnitude of a net charge within a closed surface?

The magnitude of a net charge within a closed surface is the total electric charge enclosed by the surface. It is a measure of the strength of the electric field within the surface.

2. How is the magnitude of a net charge within a closed surface calculated?

The magnitude of a net charge within a closed surface can be calculated by summing up the charges of all particles within the surface. The charges can be positive or negative, and their values are typically measured in Coulombs (C).

3. What is the significance of the magnitude of a net charge within a closed surface?

The magnitude of a net charge within a closed surface is an important quantity in understanding the behavior of electric fields. It helps determine the strength and direction of the electric field within the surface, and can also be used to calculate the electric flux through the surface.

4. Can the magnitude of a net charge within a closed surface be zero?

Yes, the magnitude of a net charge within a closed surface can be zero if the net charge enclosed by the surface is equal to zero. This can happen if the positive and negative charges within the surface cancel each other out.

5. How does the magnitude of a net charge within a closed surface affect the electric potential within the surface?

The magnitude of a net charge within a closed surface is directly proportional to the electric potential within the surface. This means that as the magnitude of the net charge increases, the electric potential also increases, and vice versa.

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