Question

A tetrahedron has an equilateral triangle base with 27.0-cm-long edges and three equilateral triangle sides. The base is parallel to the ground, and a vertical uniform electric field of strength 280 N/C passes upward through the tetrahedron.What is the electric flux through each of the three sides?

Answer 1

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:

$A = \frac{\sqrt{3}}{4}l^{2}$

$A = \frac{\sqrt{3}}{4}\times (0.27)^{2} = 0.0316\ m^{2}$

Now, the electric flux that passes through the base is given by:

$\phi = - E\cdot A = - EA$

$\phi = 280\times 0.0316 = - 8.85\ Wb$

Now, the overall flux that passes through the surface and the base of the tetrahedron is zero.

$\phi_{total} = \phi + 3phi_{surface}$

$0 = \phi + 3phi_{surface}$

$phi_{surface} = -\frac{\phi }{3}$

$phi_{surface} = -\frac{- 8.85}{3} = 2.95\ Wb$