(a)

The angle between surface 3 and the electric field is [katex]90^\circ [/katex].

Here is the electric flux through the surface 3.

[katex]{\phi _3} = \vec E(r)dA\cos \theta [/katex]

Substitute [katex]90^\circ [/katex]

[katex]\begin{array}{c}\\{\phi _3} = \vec E(r)dA\cos 90^\circ \\\\ = 0\\\end{array}[/katex]

(b)

The angle between surface 1 and the electric field is [katex]0^\circ [/katex]

Here is the electric flux through the surface 1.

[katex]{\phi _1} = {\vec E_1}(r)dA\cos \theta [/katex]

Substitute [katex]0^\circ [/katex]

[katex]\phi [/katex]1

Here is the electric field for surface 1.

[katex]\phi [/katex]2

Here is the area element for surface 1.

[katex]\phi [/katex]3

Substitute [katex]\phi [/katex]4

[katex]\phi [/katex]5

(c)

The angle between surface 2 and the electric field is [katex]0^\circ [/katex]

Here is the electric flux through surface 2:

[katex]\phi [/katex]7

Substitute [katex]0^\circ [/katex]

[katex]\phi [/katex]9

Here is the electric field for surface 2.

[katex]\begin{array}{c}\\\phi = E \cdot dA\\\\ = EdA\cos \theta \\\end{array}[/katex]0

Here is the area element for surface 2.

[katex]\begin{array}{c}\\\phi = E \cdot dA\\\\ = EdA\cos \theta \\\end{array}[/katex]1

Substitute [katex]\begin{array}{c}\\\phi = E \cdot dA\\\\ = EdA\cos \theta \\\end{array}[/katex]2

[katex]\begin{array}{c}\\\phi = E \cdot dA\\\\ = EdA\cos \theta \\\end{array}[/katex]3Ans Part a

Part of

Answer

The electric flux through the surface 3 equals 0.

Part b

[katex]\begin{array}{c}\\\phi = E \cdot dA\\\\ = EdA\cos \theta \\\end{array}[/katex]4Part C is the electric flux through the surface 1.

[katex]\begin{array}{c}\\\phi = E \cdot dA\\\\ = EdA\cos \theta \\\end{array}[/katex]4 is the electric flux through surface 2.