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2.A: Gauss's Law (Answers)
https://phys.libretexts.org/Courses/Muhlenberg_College/Physics_122%3A_General_Physics_II_(Collett)/02%3A_Gauss's_Law/2.A%3A_Gauss's_Law_(Answers)
The shaded side of the small cube would be 1/24th of the total area of the large cube; therefore, the flux through the shaded area would be $\displaystyle Φ=\frac{1}{24}\frac{q}{ε_0}$ . 77. \(\displa...The shaded side of the small cube would be 1/24th of the total area of the large cube; therefore, the flux through the shaded area would be $\displaystyle Φ=\frac{1}{24}\frac{q}{ε_0}$ . 77. $\displaystyle Φ=\frac{q_{enc}}{ε_0}$ ; There are two contributions to the surface integral: one at the side of the rectangle at $\displaystyle x=0$ and the other at the side at $\displaystyle x=2.0m$ ; $\displaystyle −E(0)[1.5m^2]+E(2.0m)[1.5m^2]=\frac{q_{enc}}{ε_0}=−100Nm2^/C$
17.8: Gauss's Law (Answers)
https://phys.libretexts.org/Courses/Kettering_University/Electricity_and_Magnetism_with_Applications_to_Amateur_Radio_and_Wireless_Technology/17%3A_Gauss's_Law_for_Calculation_of_Electrical_Field_from_Charge_Distributions/17.08%3A_Gauss's_Law_(Answers)
The shaded side of the small cube would be 1/24th of the total area of the large cube; therefore, the flux through the shaded area would be $\displaystyle Φ=\frac{1}{24}\frac{q}{ε_0}$ . 77. \(\displa...The shaded side of the small cube would be 1/24th of the total area of the large cube; therefore, the flux through the shaded area would be $\displaystyle Φ=\frac{1}{24}\frac{q}{ε_0}$ . 77. $\displaystyle Φ=\frac{q_{enc}}{ε_0}$ ; There are two contributions to the surface integral: one at the side of the rectangle at $\displaystyle x=0$ and the other at the side at $\displaystyle x=2.0m$ ; $\displaystyle −E(0)[1.5m^2]+E(2.0m)[1.5m^2]=\frac{q_{enc}}{ε_0}=−100Nm2^/C$
8: Gauss's Law (Answers)
https://phys.libretexts.org/Courses/Joliet_Junior_College/PHYS202_-_JJC_-_Testing/06%3A_Gauss's_Law/08%3A_Gauss's_Law_(Answers)
The shaded side of the small cube would be 1/24th of the total area of the large cube; therefore, the flux through the shaded area would be $\displaystyle Φ=\frac{1}{24}\frac{q}{ε_0}$ . 77. \(\displa...The shaded side of the small cube would be 1/24th of the total area of the large cube; therefore, the flux through the shaded area would be $\displaystyle Φ=\frac{1}{24}\frac{q}{ε_0}$ . 77. $\displaystyle Φ=\frac{q_{enc}}{ε_0}$ ; There are two contributions to the surface integral: one at the side of the rectangle at $\displaystyle x=0$ and the other at the side at $\displaystyle x=2.0m$ ; $\displaystyle −E(0)[1.5m^2]+E(2.0m)[1.5m^2]=\frac{q_{enc}}{ε_0}=−100Nm2^/C$

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