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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![### 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F759a3292-35c4-4e1d-95e4-a498e23761c4%2F9c8fc442-b621-4cfc-8bef-9b11677b4ca8%2Fmgu6ph5_processed.png&w=3840&q=75)
Transcribed Image Text:### 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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