How to Calculate Electric Flux through a Pyramid's Slanted Surfaces?

AI Thread Summary
To calculate the electric flux through the slanted surfaces of a pyramid with a square base in a vertical electric field, the formula Phi = E*A*cos(theta) is used. The pyramid has a base of 6.00 m and a height of 4.00 m, with an electric field strength of 52.0 N/C. The challenge lies in determining the angle between the area vector of the slanted surfaces and the electric field vector. A hint suggests considering the relationship between the flux through the slanted surfaces and the base, likening flux to flowing water to understand the net flow. Understanding these relationships is crucial for solving the problem effectively.
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Homework Statement


A pyramid with horizontal square base, 6.00 m on each side, and a hieght of 4.00 m is placed in a vertical electric field of 52.0 N/C. Calculate the total electric flux through the pyramid's four slanted surfaces.


Homework Equations



Phi = E*A*cos(theta)

The Attempt at a Solution

 
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shinobi12 said:
A pyramid with horizontal square base, 6.00 m on each side, and a hieght of 4.00 m is placed in a vertical electric field of 52.0 N/C. Calculate the total electric flux through the pyramid's four slanted surfaces.

Hi shinobi12! :smile:

Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:
 
i found the length of the of the hypotenuse and found the area of the side and I am having trouble figuring out the angle of the area vector and the electric field vector
 
shinobi12 said:
i found the length of the of the hypotenuse and found the area of the side and I am having trouble figuring out the angle of the area vector and the electric field vector

Hint: what is the relationship between the flux through the four slanted surfaces, and the flux through the base? :wink:
 
think of flux as flowing water. what is the net flow of water through the walls of the pyramid?
 
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