Flux due to a point charge

In summary, the problem involves calculating the total electric flux through a spherical surface centered at the origin and with a given radius, due to two point charges located on the x-axis and y-axis. The solution involves using Gauss's law, which states that the flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space. Through this method, the electric flux is found to be -1992.28 Nm^2/C.
  • #1
Black Armadillo
12
0

Homework Statement


A point charge q_1 = 3.45 nC is located on the x-axis at x = 1.90 m, and a second point charge q_2 = -6.95 nC is on the y-axis at y = 1.20 m.

What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius r_2 = 1.65 m?


Homework Equations


[tex]\Phi=\oint E_\bot dA[/tex]
[tex]A=4\pi r^2[/tex]
[tex]E=\frac{kq}{r^2}[/tex]


The Attempt at a Solution


I started with:
[tex]\Phi=\oint \frac{kq}{r^2} dA[/tex]
[tex]\Phi=A \oint \frac{kq}{r^2} dy[/tex]
[tex]\Phi=4\pi r^2 \oint \frac{kq}{r^2} dy[/tex]

To get r I did:
[tex]x^2+y^2=r^2[/tex]
[tex]x^2+y^2=1.65^2[/tex]
[tex]x=\sqrt{1.65^2-y^2}[/tex]

[tex]r=\sqrt{(\sqrt{1.65^2-y^2})^2+(y-1.20)^2}[/tex]

So:
[tex]\Phi=4\pi r^2 \oint \frac{kq}{1.65^2-y^2+(y-1.20)^2} dy[/tex]

Evaluating this integral from -1.65 to 1.65 gives -1992.28 Nm^2/C

I'm pretty sure I'm setting up this integral completely wrong. Any help on how to do it correctly would be greatly appreciated. Thanks in advanced for your help.
 
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  • #2
Why mess around with integrals? Use Gauss's law.
 
  • #3
remember that the flux through a closed surface is equal to the charge enclosed divided by epsilon_not. aka gauss's law
 
  • #4
By use Gauss's law do you mean [tex]\Phi=\frac{q}{\epsilon_0}[/tex]? If so don't I need to know that permittivity of free space ([tex]\epsilon_0[/tex]), which isn't given in the problem?
 
  • #5
Alright I found that epsilon_0 = 8.854E-12. Thanks for your help.
 

1. What is the concept of flux due to a point charge?

The concept of flux due to a point charge is the measure of the electric field passing through a given surface around the point charge. It represents the total number of electric field lines originating from the point charge and passing through the surface.

2. How is flux due to a point charge calculated?

The flux due to a point charge is calculated by multiplying the electric field strength at a given point by the area of the surface that the electric field lines are passing through. This can be represented mathematically as Φ = E * A, where Φ is the flux, E is the electric field strength, and A is the area of the surface.

3. What is the unit of measurement for flux due to a point charge?

The unit of measurement for flux due to a point charge is volts (V) per meter squared (m^2) or newtons (N) per coulomb (C).

4. What is the difference between electric flux and magnetic flux?

Electric flux and magnetic flux are both measures of the flow of electric and magnetic fields, respectively. The main difference between them is that electric flux is caused by an electric field, while magnetic flux is caused by a magnetic field.

5. How does the distance from the point charge affect the flux?

The distance from the point charge affects the flux as it follows an inverse square law relationship. This means that as the distance increases, the flux decreases. This is because the electric field strength decreases as the distance from the point charge increases.

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