Solving for Flux Using Enclosed Charge

In summary, the conversation discusses finding the total flux passing through the surface of a cylinder with two line charges present. The electric field is not constant on the surface, so using the formula AE does not work. Instead, the total flux is equal to the enclosed charge divided by ε0, which can be calculated using the given charge densities and length. The method for finding the enclosed charge does not depend on the type of field being dealt with.
  • #1
InertialRef
25
0

Homework Statement



https://www.smartphysics.com/Content/Media/Images/EM/03/h3_lineD.png

Where, λ1 = -3.2 μC/cm, λ2 = 9.6 μC/cm, a = 8.5 cm and h = 11.4 cm. a/2 = 4.25 cm.

What is the total flux that passes through the surface of the cylinder?

Homework Equations



E = 2kλ/r
[itex]\Phi[/itex] = E*A = qenclosed/[itex]\epsilon[/itex]o

The Attempt at a Solution



The first thing I did was solve for the total electric field of the two line charges, which is 338823529.411765 N/C. After this, the next step would be to simply multiply by the surface area of the cylinder. However, this doesn't work because the electric field is not constant throughout the surface of the cylinder.

My next step was to try and solve it using the enclosed charge, but since I have only been given the charge densities of each line, I have no clue how to figure out enclosed charge. Is there a specific method as to how you can do this?

Any help is greatly appreciated.
 
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  • #2
InertialRef said:

Homework Statement



https://www.smartphysics.com/Content/Media/Images/EM/03/h3_lineD.png

Where, λ1 = -3.2 μC/cm, λ2 = 9.6 μC/cm, a = 8.5 cm and h = 11.4 cm. a/2 = 4.25 cm.

What is the total flux that passes through the surface of the cylinder?

Homework Equations



E = 2kλ/r
[itex]\Phi[/itex] = E*A = qenclosed/[itex]\epsilon[/itex]o

The Attempt at a Solution



The first thing I did was solve for the total electric field of the two line charges, which is 338823529.411765 N/C. After this, the next step would be to simply multiply by the surface area of the cylinder. However, this doesn't work because the electric field is not constant throughout the surface of the cylinder.

My next step was to try and solve it using the enclosed charge, but since I have only been given the charge densities of each line, I have no clue how to figure out enclosed charge. Is there a specific method as to how you can do this?
Any help is greatly appreciated.

As you have noted, the electric field is not constant on the surface, so it is not AE. But the total flux is always equal to the enclosed charge divided by ε0: you need to calculate that only from the λ-s and length h.
 
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  • #3
ehild said:
As you have noted, the electric field is not constant on the surface, so it is not AE. But the total flux is always equal to the enclosed charge divided by ε0: you need to calculate that only from the λ-s and length h.

Alright, thank you. :) I think I got it now.

I do have one more question though. Is the method used to find the enclosed charge different depending upon what type of field you are dealing with?
 
  • #4
I do not get you. It is valid for the relation between electric flux and enclosed charge.

ehild
 
  • #5




To solve for the total flux passing through the surface of the cylinder, we can use the formula \Phi = E*A, where E is the electric field and A is the surface area of the cylinder. However, as you have correctly pointed out, the electric field is not constant throughout the surface of the cylinder, so we cannot simply multiply by the surface area.

To solve this problem, we can use the concept of enclosed charge. Enclosed charge is the total charge that is enclosed within a surface. In this case, the surface is the cylinder and the enclosed charge is the total charge of both line charges that are contained within the cylinder.

To find the enclosed charge, we need to calculate the length of each line charge that is within the cylinder. This can be done by using the formula for the length of a line segment, L = \sqrt{(h-a)^2 + (a/2)^2}. In this case, L1 = 4.25 cm and L2 = 7.15 cm.

Next, we can calculate the total charge enclosed by multiplying the length of each line charge by its respective charge density. This gives us a total enclosed charge of -27.2 μC.

Now, we can use the formula \Phi = qenclosed/\epsilono to calculate the total flux passing through the surface of the cylinder. Plugging in the value for the enclosed charge and the permittivity of free space, we get a total flux of -6.8 x 10^6 Nm^2/C.

Therefore, the total flux passing through the surface of the cylinder is -6.8 x 10^6 Nm^2/C.
 

FAQ: Solving for Flux Using Enclosed Charge

1. What is flux?

Flux is a measure of the flow of a physical quantity through a surface. In the context of electrostatics, it refers to the flow of electric field through a surface.

2. How is flux calculated?

The flux through a surface is calculated by taking the dot product of the electric field and the surface area vector at each point on the surface and then integrating over the entire surface.

3. What is enclosed charge?

Enclosed charge refers to the net charge contained within a closed surface. It can be calculated by adding up the individual charges within the surface or by 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.

4. How do you solve for flux using enclosed charge?

To solve for flux using enclosed charge, you can use the formula: Φ = Q/ε₀, where Φ is the flux, Q is the enclosed charge, and ε₀ is the permittivity of free space. Alternatively, you can also use the integral form of Gauss's law to calculate the flux.

5. What are some applications of solving for flux using enclosed charge?

Solving for flux using enclosed charge is useful in various areas of physics, such as electrostatics, electromagnetism, and fluid dynamics. It is also commonly used in engineering, particularly in designing electronic circuits and studying the behavior of electric fields in different materials.

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