Question

A tetrahedron has an equilateral triangle base with$20-cm$-long edges and three equilateral triangle sides. The base is parallel to the ground, and a vertical uniform electric field of strength $200N/C$ passes upward through the tetrahedron. a. What is the electric flux through the base? b. What is the electric flux through each of the three sides?

Short answer

1. The current flowing through the base is$3.464{\mathrm{Nm}}^2/\mathrm{C}$.
2. Each of the three sides has a different electric flux$1.155{\mathrm{Nm}}^2/\mathrm{C}$.

Step-by-step solution

Part (a) Step 1. Given information

Each of the equilateral triangle's sides is$20\mathrm{cm}$ in length, and the electric field strength is$200\mathrm{N}/\mathrm{C}$.

Part (a) Step 2. Find the electric flux thorough base

Consider that the electric flux ${\Phi }_E=\overrightarrow{E}·\overrightarrow{A}$

For the base, $\overrightarrow{E}$ and $\overrightarrow{A}$ are opposite each other, since$\overrightarrow{E}$ is upward through the tetrahedron

$\begin{array}{rcl}\therefore {\Phi }_E& =& EA\cos 180°\\ & =& -EA\end{array}$

where, $E=200\mathrm{N}/\mathrm{C}$

A= area of the equilateral triangle with side $a=20\mathrm{cm}$

$$\begin{gathered} =\frac{\sqrt{3}}{4}{a}^2 \\ =\frac{\sqrt{3}}{4}{\left(20\times {10}^{-2}\mathrm{m}\right)}^2 \\ =0.01732{\mathrm{m}}^2 \end{gathered}$$

$\begin{array}{rcl}\therefore {\Phi }_E& =& -(200\mathrm{N}/\mathrm{C})\left(0.01732{\mathrm{m}}^2\right)\\ & =& -3.464{\mathrm{Nm}}^2/\mathrm{C}\end{array}$

Part (b) Step 1. Given information

Each of the equilateral triangle's sides is$20\mathrm{cm}$ in length, and the electric field strength is $200\mathrm{N}/\mathrm{C}$.

Part (b) Step 2.  Find the each of the three sides has its own electric flux.

First, we must determine the angle between$\overrightarrow{E}$and$\overrightarrow{A}$ the sides of the tetrahedron.

This is equal to the angle formed by any two tetrahedron faces. A tetrahedron's angle between any two faces, on the other hand, is${\cos }^{-1}\left(\frac{1}{3}\right)$

$\begin{array}{rcl}\therefore {\Phi }_E& =& EA\cos \left[{\cos }^{-1}\left(\frac{1}{3}\right)\right]\\ & =& (200\mathrm{N}/\mathrm{C})\left(0.01732{\mathrm{m}}^2\right)\left(\frac{1}{3}\right)\\ & =& 1.155{\mathrm{Nm}}^2/\mathrm{C}\end{array}$