Answer:
The electric flux through each of the 3 sides is 2.95 Wb
Solution:
As per the question:
Length of the edges, l = 27.0 cm = 0.27 m
Strength of the electric field, E = 280 N/C
To calculate the electric field through each of the three sides:
Area of the equilateral triangle is given by:
[tex]A = \frac{\sqrt{3}}{4}l^{2}[/tex]
[tex]A = \frac{\sqrt{3}}{4}\times (0.27)^{2} = 0.0316\ m^{2}[/tex]
Now, the electric flux that passes through the base is given by:
[tex]\phi = - E\cdot A = - EA[/tex]
[tex]\phi = 280\times 0.0316 = - 8.85\ Wb[/tex]
Now, the overall flux that passes through the surface and the base of the tetrahedron is zero.
[tex]\phi_{total} = \phi + 3phi_{surface}[/tex]
[tex]0 = \phi + 3phi_{surface}[/tex]
[tex]phi_{surface} = -\frac{\phi }{3}[/tex]
[tex]phi_{surface} = -\frac{- 8.85}{3} = 2.95\ Wb[/tex]
