Flux through surface of half a cone

In summary, the problem involves finding the electric flux entering the left hand side of a cone with base radius R and height h, where the electric field penetrates the cone horizontally on a horizontal table. The formula used is the surface integral of EdA, where E represents the component of the electric field normal to the surface. The attempt at a solution involves finding the integral of Ecos(theta)dA and integrating with respect to A, resulting in a final answer of 2piERh. However, the correct answer is ERh, as the angle between E and A is not always 90 degrees and needs to be considered when finding the effective surface area facing the flux.
  • #1
physgirl
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Homework Statement


find electric flux that enters left hand side of a cone (with base radius R and height h). electric field penetrates the cone horizontally (cone is on a horizontal table).


Homework Equations


flux = surface integral of EdA
where E represents component of electric field normal to surface


The Attempt at a Solution


so that's the formula I got from my book... so I thought I had to find the integral of Ecos(theta)dA... since Ecos(theta) would be the component of E that's normal to slanted surface of the cone... and if I integrate with respect to A I get Ecos(theta)A... cos(theta) is h/sqrt(h^2+R^2)... area of cone (minus the base) is pi*R*sqrt(h^2+R^2)...

so I did... E*(h/sqrt(h^2+R^2))*[pi*R*sqrt(h^2+R^2)]/2
once I cancel out stuff, I get: E*(h)(pi*R)/2... 2piERh
however, the back of my book says the answer is ERh... why??
 
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  • #2
If theta is the semi-vertical angle (as it appears to be), then that is not the angle between the E field and the normal to each area element (it will only be true of those field lines that pass through the axis of the cone). Consider, for instance, a field line that just barely glances past the surface (i.e., is tangential to it) - clearly the angle between E and A is then 90*.

You need to first be clear about the angles involved before you can proceed. Does this make sense to you?

PS: Hint for proceeding - notice the relation between the dot product of two vectors and the projection of one vector along the other.
 
Last edited:
  • #3
the reason why your ans is wrong is because the angle that you've calculated need some thinking, and think of the effective surface area facing the flux, knowing that, you should have your answer
 

1. What is flux through the surface of half a cone?

Flux through the surface of half a cone refers to the amount of fluid, such as air or water, passing through the curved surface of a half cone in a given amount of time.

2. How is flux through the surface of half a cone calculated?

The flux through the surface of half a cone is calculated using the formula F = ρAv, where ρ is the density of the fluid, A is the surface area of the cone, and v is the velocity of the fluid.

3. What factors can affect the flux through the surface of half a cone?

The flux through the surface of half a cone can be affected by the density and velocity of the fluid, as well as the size and shape of the cone.

4. What is the significance of calculating flux through the surface of half a cone?

Calculating the flux through the surface of half a cone can help scientists understand the flow of fluids in various situations, such as in pipes, turbines, or other curved surfaces. This information can be used in engineering and design applications to optimize fluid flow.

5. Are there any real-world applications of flux through the surface of half a cone?

Yes, there are many real-world applications of flux through the surface of half a cone. For example, it can be used in the design of aircraft wings to optimize aerodynamics, or in the design of wind turbines to maximize energy production. It is also used in the study of ocean currents and weather patterns.

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