Calculating Electric Flux Through a Cone

In summary, the electric flux through the sloping side surface area of the cone is zero due to the uniform electric field.
  • #1
forestmine
203
0

Homework Statement



A cone is resting on a tabletop as shown in the figure with its face horizontal. A uniform electric field of magnitude 4550 N/C points vertically upward. How much electric flux passes through the sloping side surface area of the cone?

p18.2.jpg


Homework Equations



flux = ∫ E*dA

The Attempt at a Solution



I'm really just confused given the geometry of the situation. In general, I understand that to find the electric flux, you choose a Gaussian surface (exploiting symmetries when possible), and the flux will be equal to the electric field times the area of the Gaussian surface, which is equal to the charge enclosed divided by ε_0.

But what I'm unsure about is what kind of Gaussian surface to use here?

As for finding the charge, since the electric field is uniform and vertically upward, should I treat it as a line of charge?

This cone business is confusing me!

Any help would be greatly appreciated.
Thank you!
 
Physics news on Phys.org
  • #2
well you can treat cone itself as the gaussian surface. Since there is only constant electric
field, no charges are present inside the cone. so by gauss's law, total flux is zero. but the total flux is flux through the slanted surface + the flux through the flat surface. since E points vertically upwards, its easy to calculate the flux through the flat surface. that will then give you the flux through the slanted surface.
 
  • #3
Turns out the total flux is not 0, so I'm assuming the charge that emits the electric field is to be enclosed in the gaussian surface, even though they don't mention any such charge. And so that's where I'm confused...how do I find the charge. From that point, finding the flux should be easy enough, but as it is, I don't have a q(enc). Also, since E is constant and vertically upwards, then I'm assuming it comes from a line of charge, so maybe I ought to be using lamba as line charge density, and use q=(lamba)l, but these are all unknown values. I don't see how to compute this gives what I have?
 
  • #4
well this constant E is created by charges which are far away. so don't worry about them. at the location of this cone, you just have a constant E. problem is simple
 
  • #5
Ok, wow, so mathematically, the problem is very simple. Thank you!

However, I don't understand why there is a flux. I thought that in order for there to be a flux, there must be a charge enclosed...in this case there's not.
 
  • #6
the net flux through the entire cone is zero. its like water flowing through some closed imaginary surface in the river. if we only look at some part of the surface, then there is water going through it. But on the whole, the net water flowing out is zero, since water coming in is same as water going out
 
  • #7
I see. So, attention to detail is key, here.

Thank you so much for the help! I really appreciate it, Mr. Newton, as well as all of your laws. ;)
 

What is electric flux through a cone?

Electric flux through a cone is the measure of the electric field passing through a conical surface. It is a measure of the total number of electric field lines passing through the surface.

How is electric flux through a cone calculated?

The electric flux through a cone can be calculated by taking the dot product of the electric field and the area vector of the cone's surface. Mathematically, it is expressed as Φ = E · A · cosθ, where E is the electric field, A is the surface area of the cone, and θ is the angle between the electric field and the area vector.

What factors affect the electric flux through a cone?

The electric flux through a cone is affected by the strength of the electric field, the surface area of the cone, and the angle between the electric field and the area vector. It is also affected by the shape and orientation of the cone.

What is the unit of measurement for electric flux through a cone?

The unit of measurement for electric flux through a cone is volt-meters (V·m) or newton-meters squared per coulomb (N·m²/C). This unit is derived from the basic unit of electric field, volts per meter (V/m).

What are the applications of electric flux through a cone?

Electric flux through a cone is used in calculating the electric field strength within a cone-shaped capacitor, which is used in electronic devices. It is also used in measuring the electric field produced by a point charge or a line charge, which can be useful in determining the electric field around a charged object. Additionally, the concept of electric flux through a cone is important in understanding and solving problems related to electromagnetism and electric circuits.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
902
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
679
  • Introductory Physics Homework Help
Replies
11
Views
399
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
3K
Back
Top