A tetrahedron contains a point charge Q = 65 nC. Find the electric flux through
ID: 2144562 • Letter: A
Question
A tetrahedron contains a point charge Q = 65 nC. Find the electric flux through the surface of the tetrahedrom. Find the flux through each face of the tetrahedron if the charge is at the center of the tetrahedron.
A cube of side l = 2.0 m is centered at the origin, with the coordinate axes perpendicular to its faces. Find the flux of the electric field E = (15,27,39) N/C through each face of the cube.
Find the flux of the electric field E = 250 N/C in the y direction through the surface x2 + y2 + z2 = 1.0 m2, x > 0, y > 0, z > 0.
The electric field in a certain region of space has components Ex = 6.0 N/C, Ey = 7.0 N/C, and Ez = 0. Find the electric flux though the surface x = 6y, 0 < x < 6.0 m, 0 < z < 1.0 m. What is the angle between the electric field and the unit vector normal to the surface?
I would appreciate more than just plain answers. I am behind on material and would like a step by step approach.
Explanation / Answer
1)
total electric flux = Q/epsilon = 7341.3 N.m^2/C
it has four faces.
fluxx through each face = total flux/4 = 1835.33 N.m^2/C
2)
electric flux through +x face = Ex*A*cos(0) = 15*4 = +60 N.m^2/c
electric flux through -x face = Ex*A*cos(theta) = 15*4*cos(180) = -60 N.m^2/c
electric flux through +y face = Ey*A*cos(theta) = 27*4*cos(0) = +108 N.m^2/c
electric flux through -y face = Ey*A*cos(theta) = 27*4*cos(180) = -108 N.m^2/c
electric flux through +z face = Ez*A*cos(theta) = 39*4*cos(0) = +156 N.m^2/c
electric flux through -z face = Ez*A*cos(theta) = 39*4*cos(1800) = -156 N.m^2/c
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