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**a) What is the electric flux ? _{3} through the annular ring, surface 3?** (Express your answer in terms of

*C*,

*r*1,

*r*2, and any constants.)

**b) What is the electric flux ? _{1} through surface 1?** (Express ?

_{1}in terms of

*C*,

*r*1,

*r*2, and any needed constants.)

**c) What is the electric flux ? _{2} passing outward through surface 2?** (Express ?

_{2}in terms of

*r*1,

*r*2,

*C*, and any constants or other known quantities.)

Concepts and Reason_{3} firstFundamentalsThis problem can be solved using the concept of electric flux.

Calculate the electric flux [katex]{s_3}[/katex] first

The electric flux can be defined as the area covered by the electric field lines. The surface area will increase if there are more field lines crossing the surface.

Mathematically, an electric flux can be described as the dot product from the electric field and the component of area perpendicular the field.

The expression of electric flux ([katex]\phi [/katex])

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

Ehere is the electric field.Athe area is the angle between the area element and the electric field.As a result of point charge

Q, the electric field at positionris shown as follows:[katex]\vec E(r) = \frac{C}{{{r^2}}}\hat r[/katex]

Chere is the constant proportional charge.(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 ofAnswer

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.