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Where ε o 885x10 12 c 2 nm 2 and is called the

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where ε o = 8.85x10 –12 C 2 /(Nm 2 ) and is called the permittivity of free space. If we try to find the electric flux through the Gaussian surface we change our integral to: where the loop on the integral sign means we take the integration over the entire closed surface.
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Gauss’ Law What if there were no charges present inside the closed Gaussian surface, what would the electric flux be? But if a net positive charge is present, then there is a source of electric field lines and the electric flux cannot be zero. The amount of field lines that enter the surface equal the amount of field lines that leave the surface. This leads to a net electric flux of zero.
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Applications of Gauss’ Law The power of Gauss’ Law is to use the net charge enclosed in a surface to find the electric field at a certain distance (or vice versa). The trick comes in finding a proper Gaussian surface. If you are looking for the electric field from a configuration of charges, you would like to choose a surface in which the electric field is constant (in magnitude) at all points on the surface. This allows you to pull E out of the integral.
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Gauss’ Law Example Use Gauss’ Law to find the electric field a distance r from a lone positive point charge +q. Answer Define a coordinate system. Choose the point charge as r = 0. Any outward distance would be positive.
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Gauss’ Law Answer If we are going to use Gauss’ Law we must choose a Gaussian surface.
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