Chapter 1: Negative Charge B

System Configuration and Geometry

• System components:

  • Positive charge $A$

  • Negative charge $B$

• Bounded regional evaluation surfaces:

  • Closed surface $A$

  • Closed surface $B$

  • Closed surface $C$

  • Closed surface $D$

Core Principles of Electric Flux and Gauss's Law

• Gauss's Law defines the relationship between total electric flux $\Phi_E$ passing through any closed surface and the net charge $Q_{\text{enc}}$ enclosed within that volume: $\Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$

  • $\Phi_E$: Net electric flux through the closed surface.

  • $\mathbf{E}$: Electric field vector.

  • $d\mathbf{A}$: Differential area vector directed orthogonally outward from the closed surface.

  • $Q_{\text{enc}}$: Net algebraic charge enclosed inside the surface boundary.

  • $\varepsilon_0$: Vacuum permittivity constant.

Electric Flux Calculations Across Closed Surfaces

• Closed Surface $A$:

  • Encloses positive charge $A$

  • Net enclosed charge: $Q_{\text{enc}} = +q_A$

  • Net flux through surface $A$ is positive (directed outward): $\Phi_{E, A} = \frac{+q_A}{\varepsilon_0}$

• Closed Surface $B$:

  • Encloses negative charge $B$

  • Net enclosed charge: $Q_{\text{enc}} = -q_B$

  • Net flux through surface $B$ is negative (directed inward): $\Phi_{E, B} = \frac{-q_B}{\varepsilon_0}$

• Closed Surface $C$:

  • Encloses both positive charge $A$ and negative charge $B$

  • Net enclosed charge: $Q_{\text{enc}} = q_A + q_B = q_A - |q_B|$

  • If charges are equal in magnitude ($|q_A| = |q_B|$), the net enclosed charge $Q_{\text{enc}} = 0$

  • Resulting electric flux through surface $C$: $\Phi_{E, C} = 0$

• Closed Surface $D$:

  • Encloses zero charges ($Q_{\text{enc}} = 0$

  • Net electric flux through surface $D$: $\Phi_{E, D} = 0$

System Configuration and Geometry

• System components:

  • Positive charge $A$

  • Negative charge $B$

• Bounded regional evaluation surfaces:

  • Closed surface $A$

  • Closed surface $B$

  • Closed surface $C$

  • Closed surface $D$

Core Principles of Electric Flux and Gauss's Law

• Gauss's Law defines the relationship between total electric flux $\ Phi<em>E$ passing through any closed surface and the net charge $Q</em>{\text{enc}}$ enclosed within that volume: $\Phi<em>E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q</em>{\text{enc}}}{\varepsilon_0}$

  • $\Phi_E$: Net electric flux through the closed surface.

  • $\mathbf{E}$: Electric field vector.

  • $d\mathbf{A}$: Differential area vector directed orthogonally outward from the closed surface.

  • $Q_{\text{enc}}$: Net algebraic charge enclosed inside the surface boundary.

  • $\varepsilon<em>0$: Vacuum permittivity constant ($\varepsilon</em>0 \approx 8.854 \times 10^{-12}\,C^2\,N^{-1}\,m^{-2}$).

Electric Flux Calculations Across Closed Surfaces

• Closed Surface $A$:

  • Encloses positive charge $A$

  • Net enclosed charge: $Q<em>{\text{enc}} = +q</em>A$

  • Net flux through surface $A$ is positive (directed outward):
    $\Phi<em>{E, A} = \frac{+q</em>A}{\varepsilon_0}$

• Closed Surface $B$:

  • Encloses negative charge $B$

  • Net enclosed charge: $Q<em>{\text{enc}} = -q</em>B$

  • Net flux through surface $B$ is negative (directed inward):
    $\Phi<em>{E, B} = \frac{-q</em>B}{\varepsilon_0}$

• Closed Surface $C$:

  • Encloses both positive charge $A$ and negative charge $B$

  • Net enclosed charge: $Q<em>{\text{enc}} = q</em>A + q<em>B = q</em>A - |q_B|$

  • If charges are equal in magnitude ($|q<em>A| = |q</em>B|$), the net enclosed charge $Q_{\text{enc}} = 0$

  • Resulting electric flux through surface $C$:
    $\Phi_{E, C} = 0$

• Closed Surface $D$:

  • Encloses zero charges ($Q_{\text{enc}} = 0$)

  • Net electric flux through surface $D$:
    $\Phi_{E, D} = 0$

Example Problems

Problem 1: Calculating Flux for a Single Enclosed Charge

• Question: A point charge $q<em>A = +5.00 \times 10^{-9}\,C$ is enclosed inside Surface $A$. Calculate the net electric flux $\Phi</em>{E, A}$ passing through Surface A$.

• Solution:

  1. Apply Gauss's Law:
     \Phi{E, A} = \frac{Q{\text{enc}}}{\varepsilon_0}$</p></li><li><p>Substitute the values$Q{\text{enc}} = +5.00 \times 10^{-9}\,C$and$\varepsilon0 = 8.854 \times 10^{-12}\,C^2\,N^{-1}\,m^{-2}$:<br>$\Phi{E, A} = \frac{5.00 \times 10^{-9}\,C}{8.854 \times 10^{-12}\,C^2\,N^{-1}\,m^{-2}}$$\Phi{E, A} \approx 5.65 \times 10^2\,N\,m^2\,C^{-1}$</p></li></ol></li></ul><h6>Problem 2: Calculating Flux for Multiple Enclosed Charges</h6><ul><li><p><strong>Question:</strong> Surface$C$encloses two charges:$qA = +3.00 \times 10^{-9}\,C$and$qB = -7.00 \times 10^{-9}\,C$. Determine the net electric flux$\Phi_{E, C}$through Surface$C$.

  2. Solution:

     1. Find the net enclosed charge $Q<em>{\text{enc}}$: $Q</em>{\text{enc}} = q<em>A + q</em>B$
        $Q_{\text{enc}} = (3.00 \times 10^{-9}\,C) + (-7.00 \times 10^{-9}\,C) = -4.00 \times 10^{-9}\,C$

     2. Calculate the electric flux:
        $\Phi<em>{E, C} = \frac{-4.00 \times 10^{-9}\,C}{8.854 \times 10^{-12}\,C^2\,N^{-1}\,m^{-2}}$ $\Phi</em>{E, C} \approx -4.52 \times 10^2\,N\,m^2\,C^{-1}$