ece106: attempt flux
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@ -221,3 +221,30 @@ $$dQ=\rho_s dS$$
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4. Create a right-angle triangle with $A$, the desired point, and usually the origin
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5. Attempt to find symmetry
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6. Solve
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## Gauss's law
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!!! definition
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- A **closed surface** is any closed three-dimensional object.
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- **Electric flux** represents the number of electric field lines going through a surface.
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At an arbitrary surface, the **normal** to the plane is its vector form:
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$$\vec{dS}=\vec n\cdot dS$$
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The **electric flux density** $\vec D$ is an alternate representation of electric field strength. In a vacuum:
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$$\vec D = \epsilon_0\vec E$$
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**Electric flux** is the electric flux density multiplied by the surface area at every point of an object.
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$$\phi_e=\epsilon_0\int_s\vec E\bullet\vec{dS}$$
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The flux from charges outside a closed surface will **always be zero at the surface**. A point charge in the centre of a closed space has a flux equal to its charge. Regardless of the charge distribution or shape, the **total flux** through a closed surface is equal to the **total charge within** the closed surface.
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$$\oint \vec D\bullet\vec{dS}=Q_\text{enclosed}$$
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This implies $\phi_e>0$ is a net positive charge enclosed.
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!!! warning
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Gauss's law only applies when $\vec E$ is from all charges in the system
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