Question

a) What is the electric flux ?₃ through the annular ring, surface 3? (Express your answer in terms of C, r1, r2, and any constants.)

b) What is the electric flux ?₁ through surface 1? (Express ?₁ in terms of C, r1, r2, and any needed constants.)

c) What is the electric flux ?₂ passing outward through surface 2? (Express ?₂ in terms of r1, r2, C, and any constants or other known quantities.)

Answer 1

Concepts and reason

The concept used to solve this problem is electric flux.

Initially, calculate the electric flux ${s_3}$ through the angular ring, surface 3. After that calculate the electric flux ${s_1}$ through the angular ring, surface 1. Finally, calculate the electric flux ${s_2}$ through the angular ring, surface 1.

Fundamentals

The electric flux is defined as the measure of the electric field lines passing through the given area. More the field lines passing through the surface more will be the surface area and vice-versa.

Mathematically, the electric flux is defined as the dot product of the electric field with the component of area perpendicular to the field.

The expression of the electric flux ($\phi$) associated with the area element dA is given as follows:

$\begin{array}{c}\\\phi = E \cdot dA\\\\ = EdA\cos \theta \\\end{array}$

Here, E is the electric field, A is the area and is the angle between the electric field and area element.

The electric field at position r due to point charge q is given as follows:

$\vec E(r) = \frac{C}{{{r^2}}}\hat r$

Here, C is the constant proportional to charge.

(a)

The angle between the electric field and the area related to surface 3 is $90^\circ$.

The electric flux through the surface 3 is given as follows:

${\phi _3} = \vec E(r)dA\cos \theta$

Substitute $90^\circ$ for $\theta$ in the expression ${\phi _3} = \vec E(r)dA\cos \theta$.

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

(b)

The angle between the electric field and the area related to surface 1 is $0^\circ$.

The electric flux through the surface 1 is given as follows:

${\phi _1} = {\vec E_1}(r)dA\cos \theta$

Substitute $0^\circ$ for $\theta$ in the expression ${\phi _1} = \vec E(r)dA\cos \theta$.

$\begin{array}{c}\\{\phi _1} = {{\vec E}_1}(r)dA\cos 0^\circ \\\\ = {{\vec E}_1}(r)dA\\\end{array}$

The electric field for surface 1 is given as follows:

${\vec E_1}(r) = \frac{C}{{{r_1}^2}}$

The area element for surface 1 is given as follows:

$dA = 2\pi {r_1}^2$

Substitute $2\pi {r_1}^2$ for $dA$ and $\frac{C}{{{r_1}^2}}$ for ${\vec E_1}(r)$ in the expression ${\phi _1} = {\vec E_1}(r)dA$.

$\begin{array}{c}\\{\phi _1} = \frac{C}{{{r_1}^2}}2\pi {r_1}^2\\\\ = 2\pi C\\\end{array}$

(c)

The angle between the electric field and the area related to surface 2 is $0^\circ$.

The electric flux through the surface 2 is given as follows:

${\phi _2} = {\vec E_2}(r)dA\cos \theta$

Substitute $0^\circ$ for $\theta$ in the expression ${\phi _2} = {\vec E_2}(r)dA\cos \theta$.

$\begin{array}{c}\\{\phi _1} = {{\vec E}_2}(r)dA\cos 0^\circ \\\\ = {{\vec E}_2}(r)dA\\\end{array}$

The electric field for surface 2 is given as follows:

${\vec E_2}(r) = \frac{C}{{{r_2}^2}}$

The area element for surface 2 is given as follows:

$dA = 2\pi {r_2}^2$

Substitute $2\pi {r_2}^2$ for $dA$ and $\frac{C}{{{r_2}^2}}$ for ${\vec E_2}(r)$ in the expression ${\phi _2} = {\vec E_2}(r)dA$.

$\begin{array}{c}\\{\phi _2} = \frac{C}{{{r_2}^2}}2\pi {r_2}^2\\\\ = 2\pi C\\\end{array}$

Ans: Part a

Part a

Answer

The electric flux through the surface 3 is 0.

Part b

The electric flux through the surface 1 is $2\pi C$.

Part c

The electric flux through the surface 2 is $2\pi C$.

Answer 2

a. Obviously zero b. 2 pi C C. Same from both surface 1 and 2.