A pyramid with horizontal square base, 6.00 m on each side, and a height 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 slanted surfaces.
Electric Flux = E dot dA
The Attempt at a Solution
I realize there is an easy way to do it, calculate the area of the base * E field and then set a correct sign, got that.
Now, I should be able to calculate this the harder but just as appropriate way. Find the area of the slanted surfaces, find the vector dA by finding the normal angle to the surface compared to the E-field. Then compute cos (theta) * A * E * 4, the 4 is because there are 4 slanted surfaces total. This is where I go crazy apparently.
Find the area of the slanted surface:
A = 0.5 * base * height (lateral surface height, not pyramid height)
Base = 6.00 m
Make a right triangle from the center of the base out perpendicular to the side of the square base, this leg has length 3.00 m. Then start at the center of the square base and make the other leg go up to the top point vertice, this has length 4.00 m.
You have a 3-4-5 triangle, so the lateral height is 5.00 m
A = 0.5 * 6.00 m * 5.00 m = 15.0 m^2
Now make a vector normal to that triangular lateral surface with magnitude A.
This vector has an angle (theta) with respect to the E feild. We find that angle by noticing that a 3-4-5 triangle has angles of 60 degrees at the 3-5 junction and 30 at the 4-5 junction. We want the 3-5 junction so (theta) = 60 degrees.
4*E dot dA = 4* EA cos (theta) = 4* EA cos 60 = 4* EA * 0.5 = 2* E * 15 m^2 = E * 30 m^2
Easy Method: -(E dot dA) = -E -(area of base) cos (0)= E * (area of base) = E * 36 m^2