Electric Flux through Cylindrical Surface

In summary, the conversation discusses using Gauss' Law to solve a problem involving an infinite line of charge with charge density of -5 μC/cm. The individual tried calculating the total enclosed charge and dividing it by epsilon naught, but it did not work. They are unsure how to use the LHS in the equation and asked for a step-by-step explanation. Another individual suggests they may have made a mistake in unit conversions. The original individual later realizes their mistake and thanks the other for their help.
  • #1
dcco
2
0
An infinite line of charge with charge density λ1 = -5 μC/cm is aligned with the y-axis as shown.

dsB0K.png


I am pretty sure you are supposed to use Gauss' Law so I tried just calculating the total enclosed charge and dividing it by epsilon naught, but it's not working. I think maybe I have to use the LHS in some way, but I really have no clue how.

If someone could give a step-by-step explanation as to how to set up the equations in a general way I would be very happy. I don't need exact numbers I just need to know how to solve this (because I have a lot of other problems that I need to solve as well and I can't seem to set any of them up properly).
 
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  • #2
dcco said:
An infinite line of charge with charge density λ1 = -5 μC/cm is aligned with the y-axis as shown.

dsB0K.png


I am pretty sure you are supposed to use Gauss' Law so I tried just calculating the total enclosed charge and dividing it by epsilon naught, but it's not working. I think maybe I have to use the LHS in some way, but I really have no clue how.
Well, that's the correct way to solve the problem. You might just be messing up the unit conversions.
If someone could give a step-by-step explanation as to how to set up the equations in a general way I would be very happy. I don't need exact numbers I just need to know how to solve this (because I have a lot of other problems that I need to solve as well and I can't seem to set any of them up properly).
 
  • #3
Oh I see, I messed up my calculations but now I fixed it and it worked this time. Thank you for your help!
 

What is electric flux through a cylindrical surface?

The electric flux through a cylindrical surface is the measure of the total number of electric field lines passing through the surface. It is a measure of the amount of electric field through a given surface.

How is electric flux through a cylindrical surface calculated?

The electric flux through a cylindrical surface can be calculated by taking the dot product of the electric field vector and the surface area vector. This can be represented by the equation Φ = E * A * cos(θ), where Φ is the electric flux, E is the electric field, A is the surface area, and θ is the angle between the electric field and the surface area vector.

What is the unit of electric flux through a cylindrical surface?

The unit of electric flux is N*m²/C, which is equivalent to volts (V).

How does the shape and orientation of a cylindrical surface affect the electric flux through it?

The electric flux through a cylindrical surface is directly proportional to the surface area of the cylinder. As the surface area increases, the electric flux also increases. The orientation of the surface also affects the electric flux, as a surface parallel to the electric field lines will have a higher flux than a surface perpendicular to the field lines.

What are some real-life applications of understanding electric flux through a cylindrical surface?

Understanding electric flux through a cylindrical surface is important in various fields such as engineering, physics, and electronics. It helps in the design and analysis of electrical circuits, electric motors, and generators. It is also vital in understanding the behavior of electric fields in materials, such as in capacitors and insulators.

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