### Electric Flux Calculation through a Spherical Closed Surface**Objective:** Calculate the net electric flux through the spherical closed surface depicted in the figure.**Figure Description:** - There are three point charges labeled \( q_1 \), \( q_2 \), and \( q_3 \).- \( q_1 \) is positioned outside the spherical surface on the left side.- Charges \( q_2 \) and \( q_3 \) are located inside the spherical surface on the right.**Charge Values:** - \( q_1 = +1.88 \, \text{nC} \)- \( q_2 = +1.01 \, \text{nC} \)- \( q_3 = -3.18 \, \text{nC} \)**Instruction:** 1. The task involves calculating the net electric flux, denoted as: - \(\text{{Net Electric Flux in }} \text{{N}} \cdot \text{{m}}^2/\text{{C}}\)**Theoretical Explanation:**- According to Gauss’s Law, the electric flux \( \Phi \) through a closed surface is proportional to the charge enclosed \( q_{\text{enc}} \) by the surface: \[ \Phi = \frac{q_{\text{enc}}}{\varepsilon_0} \] where \( \varepsilon_0 \) is the permittivity of free space.**Analysis of Charge Enclosure:**- Inside the spherical surface are \( q_2 \) and \( q_3 \).- The charge \( q_1 \) does not contribute to the electric flux through the spherical surface, since it is outside.**Steps for Calculation:**1. Determine the total enclosed charge: \[ q_{\text{enc}} = q_2 + q_3 = (+1.01 \, \text{nC}) + (-3.18 \, \text{nC}) = -2.17 \, \text{nC} \]2. Use Gauss’s Law to find the net electric flux. Convert \( q_{\text{enc}} \) to Coulombs (1 nC = \( 10^{-9} \) C): \[ q_{\text{enc}} = -2.17 \times 10^{-9

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Find the net electric flux through the spherical closed surface shown in the figure below. The two charges on the right are inside the spherical surface. (Take q₁ = +1.88 nC, 92 = +1.01 nC, and 93 = -3.18 nC.) N-m²/c O 9₁ O 93

Find the net electric flux through the spherical closed surface shown in the figure below. The two charges on the right are inside the spherical surface. (Take q₁ = +1.88 nC, 92 = +1.01 nC, and 93 = -3.18 nC.) N-m²/c O 9₁ O 93

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Chapter1: Units, Trigonometry. And Vectors

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Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

### Electric Flux Calculation through a Spherical Closed Surface

**Objective:**
Calculate the net electric flux through the spherical closed surface depicted in the figure.

**Figure Description:**
- There are three point charges labeled \( q_1 \), \( q_2 \), and \( q_3 \).
- \( q_1 \) is positioned outside the spherical surface on the left side.
- Charges \( q_2 \) and \( q_3 \) are located inside the spherical surface on the right.

**Charge Values:**
- \( q_1 = +1.88 \, \text{nC} \)
- \( q_2 = +1.01 \, \text{nC} \)
- \( q_3 = -3.18 \, \text{nC} \)

**Instruction:**
1. The task involves calculating the net electric flux, denoted as:
- \(\text{{Net Electric Flux in }} \text{{N}} \cdot \text{{m}}^2/\text{{C}}\)

**Theoretical Explanation:**
- According to Gauss’s Law, the electric flux \( \Phi \) through a closed surface is proportional to the charge enclosed \( q_{\text{enc}} \) by the surface:
\[
\Phi = \frac{q_{\text{enc}}}{\varepsilon_0}
\]
where \( \varepsilon_0 \) is the permittivity of free space.

**Analysis of Charge Enclosure:**
- Inside the spherical surface are \( q_2 \) and \( q_3 \).
- The charge \( q_1 \) does not contribute to the electric flux through the spherical surface, since it is outside.

**Steps for Calculation:**
1. Determine the total enclosed charge:
\[
q_{\text{enc}} = q_2 + q_3 = (+1.01 \, \text{nC}) + (-3.18 \, \text{nC}) = -2.17 \, \text{nC}
\]
2. Use Gauss’s Law to find the net electric flux. Convert \( q_{\text{enc}} \) to Coulombs (1 nC = \( 10^{-9} \) C):
\[
q_{\text{enc}} = -2.17 \times 10^{-9

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