Determining Electric Flux for cylindrical surfaces

In summary: N/C.In summary, the problem involves calculating the total electric flux passing through a closed surface with varying surface charge densities on three cylindrical surfaces. The equation for calculating flux is given, but the method for using it is unclear. Another approach is attempted, but it is not clear if it is correct. For the second part of the question, the formula for calculating \vec{D} is used to determine the value at a specific point.
  • #1
euphtone06
22
0

Homework Statement


Cylin. surfaces [tex]\rho[/tex] = 1 , 2 , 3, cm have uniform surface charge density of 20, -8, 5 nC/m^2.

What is the total electric flux that passes through the closed surface [tex]rho[/tex] = 4 cm and z(from 0 to 1 m)?

And what is [tex]\vec{D}[/tex] at the point [tex]\rho[/tex]= 4 cm , [tex]\phi[/tex]= 0 , z = .5 cm


Homework Equations


[tex]\Phi[/tex]= E2piRL=[tex]\lambda[/tex]L/8.85x10-12
E = [tex]\rho[/tex]/(2pi8.85x10^-12 [tex]\rho[/tex])


The Attempt at a Solution



So the closed surface from which to determine the flux is at [tex]\rho[/tex] = .04 m

I frankly am unsure how to use the equations to get the total flux.

[tex]\Phi[/tex]1 = 20(.01)/8.85x10^-12
[tex]\Phi[/tex]2= -8(.02)/8.85x10^-12
[tex]\Phi[/tex]3= 5(.03)/8.85x10^-12


Is it as simple as adding them all up? Or does the center most cylindrical surface at 1 cm have an affect on 2 cm surface which then in turn affects the 3 cm? I have a feeling this is completely wrong so I attempted it a different way

[tex]\int[/tex].04*20
[tex]\int[/tex].04*-8
[tex]\int[/tex].04*5
total flux = sum of the integrals?

I obviously am struggling with this concept any help would be appreciated Thanks!
 
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  • #2
For the second part of the question what is \vec{D} at the point \rho= 4 cm , \phi= 0 , z = .5 cmI believe this is \vec{D} = (20+ -8 + 5)/2pi(4x10^-2)(8.85x10^-12)
 
  • #3




Determining the electric flux through cylindrical surfaces can be done by using the equation \Phi = \int_S \vec{E} \cdot d\vec{A}, where \vec{E} is the electric field and d\vec{A} is the differential area element on the surface. In this case, the closed surface is at \rho = 4 cm and z is varying from 0 to 1 m. To calculate the total electric flux, we need to integrate the electric field over the entire surface.

First, we need to determine the electric field at the point \rho = 4 cm, \phi = 0, z = 0.5 cm. We can use the equation \vec{E} = \frac{\rho}{2\pi\epsilon_0\rho}, where \rho is the surface charge density. Plugging in the values for \rho and \epsilon_0, we get \vec{E} = 0.0226 N/C.

Next, we can calculate the electric flux through each of the cylindrical surfaces using the equation \Phi = \int_S \vec{E} \cdot d\vec{A}. For the first surface at \rho = 1 cm, we have \Phi_1 = \int_0^{0.5} 0.0226 \cdot 2\pi(0.01)(0.01) dz = 0.000226 Nm^2/C. Similarly, for the second surface at \rho = 2 cm, we have \Phi_2 = \int_0^{0.5} 0.0226 \cdot 2\pi(0.02)(0.02) dz = 0.000905 Nm^2/C. For the third surface at \rho = 3 cm, we have \Phi_3 = \int_0^{0.5} 0.0226 \cdot 2\pi(0.03)(0.03) dz = 0.002035 Nm^2/C.

To get the total electric flux, we can simply add these values together, giving us a total flux of 0.003166 Nm^2/C. This means that 0.003166 N of electric flux passes through the closed surface at \rho = 4 cm and z varying from 0 to 1 m.

As for the electric displacement
 

1. What is Electric Flux and why is it important?

Electric Flux is a measure of the flow of an electric field through a given surface. It is important because it helps us understand the strength and direction of an electric field, which is crucial in many practical applications such as designing electronics and understanding electromagnetic phenomena.

2. How is Electric Flux calculated for cylindrical surfaces?

To calculate Electric Flux for a cylindrical surface, we use the formula Φ = EAcosθ, where Φ is the electric flux, E is the electric field strength, A is the area of the cylindrical surface, and θ is the angle between the electric field and the normal vector of the surface.

3. Can the Electric Flux be negative?

Yes, the Electric Flux can be negative. This indicates that the electric field is passing through the surface in the opposite direction of the normal vector. It is important to pay attention to the direction of the electric field and the normal vector when calculating Electric Flux.

4. What are the units of Electric Flux?

The units of Electric Flux are Nm²/C or Vm.

5. What are some real-life applications of determining Electric Flux for cylindrical surfaces?

Determining Electric Flux for cylindrical surfaces is used in various applications such as calculating the capacitance of a cylindrical capacitor, designing and optimizing electric motors, and understanding the behavior of electric fields in cylindrical conductors.

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