Electric flux through a surface

In summary: The field is the same at every point on the surface and is directed perpendicularly to the surface. So the dot product of E and dA is just EdA. We can pull E out of the integral. The integral of dA is just the area of the surface, which is d^2. So the total flux is E*d^2. The flux through a closed surface is equal to the net charge enclosed by the surface divided by the permittivity of free space (ε0). So we have:E*d^2 = qenc/ε0Since the only charge enclosed by the surface is q, we have:E*d^2 = q/ε0Solving for E, we get:E
  • #1
Plasmosis1
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Homework Statement


A point charge q is situated a distance d/2 above a square surface of side d as shown (see attached).
a) It cannot be determined
b) 2q/ε0d
c) q/4[itex]\pi[/itex]ε0
d) q/[itex]\pi[/itex]ε0d2
e) q/6ε0

Homework Equations



[itex]\Phi[/itex]=qenc0=[itex]\oint[/itex]E*dA

The Attempt at a Solution


Shouldn't the answer be zero because whatever enters the area also leave the area? If not I know I should use the integral of E*dA but I don't know what E would be.
 

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  • #2
If the area were a thin volume you'd be correct, the net flux into the volume = 0, but flux thru a surface A = the integral over the surface of the dot-product of the electric field E and an element of surface area dA: flux = ∫∫ E*dA.


So yes, you need to integrate E*dA over the surface. dA is normal to the surface at every point on the surface.
 
  • #3
You can manage without integration.
Imagine the charge is in the middle of a cube with side d.
Due to symmetry, the flux through each face of the cube is the same.
And the total flux through the cubic, closed surface can be found from Gauss' law.
 
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  • #4
Nasu has come up with the right and simple way to solve this.
 
  • #5


The correct answer is b) 2q/ε0d. This can be determined by using the equation for electric flux, Φ=qenc/ε0, where q is the point charge, enc is the enclosed charge, and ε0 is the permittivity of free space. In this case, the enclosed charge is q, and the surface area is d^2. Therefore, Φ=q/d^2*ε0. Since the charge is situated a distance d/2 above the surface, only half of the flux will pass through the surface, giving a final answer of 2q/ε0d.
 

1. What is electric flux through a surface?

Electric flux through a surface is a measure of the electric field passing through a given surface. It is represented by the symbol Φ and is defined as the product of the electric field and the area of the surface, or Φ = E∙A.

2. How is electric flux through a surface calculated?

To calculate the electric flux through a surface, you need to first determine the electric field at each point on the surface. Then, you can use the formula Φ = E∙A, where E is the magnitude of the electric field and A is the area of the surface. Finally, you can sum up the electric flux at each point on the surface to get the total electric flux through the surface.

3. What is the unit of electric flux through a surface?

The unit of electric flux through a surface is Newtons per meter squared (N∙m2) in the SI system of units. In other systems of units, it may be expressed as volts per meter (V∙m) or coulombs per meter squared (C∙m2).

4. What factors affect the electric flux through a surface?

The electric flux through a surface is affected by the strength of the electric field, the orientation of the surface with respect to the electric field, and the size of the surface. Generally, a stronger electric field or a larger surface area will result in a higher electric flux through the surface.

5. What is the significance of electric flux through a surface?

The electric flux through a surface is an important concept in understanding the behavior of electric fields. It helps us to visualize and quantify the flow of electric field lines through a given surface, and is used in many applications such as Gauss's Law and the calculation of electric potential.

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