Solve Gauss's Law and Flux: 55.3M Excess Electrons

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SUMMARY

The discussion focuses on calculating the net electric flux through a closed surface containing 55.3 million excess electrons using Gauss's Law. The relevant equation is \(\Phi = \frac{Q_{in}}{\epsilon}\), where \(Q_{in}\) is the enclosed charge and \(\epsilon\) is the permittivity constant (8.85 x 10^-12 C^2/N·m^2). The charge of a single electron is -1.60 x 10^-19 C, allowing for the calculation of the total charge from the number of excess electrons. The solution confirms that the electric field is negative due to the presence of excess electrons, leading to a straightforward application of Gauss's Law.

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dals2002
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[SOLVED] Gauss's Law and Flux

ok so i have the biggest question of all i tried solving the following problem:

55.3 million excess electrons are inside a closed surface. What is the net electric flux through the surface?

now i tried the following equations:

[tex]\Phi[/tex]= E . A = Qin/[tex]\epsilon[/tex]

but i have the biggest question when there is excess electron does it mean that the electron field is negative, and if it does then that goes in Qin where then [tex]\Phi[/tex] would be equal to Qin/[tex]\epsilon[/tex], and epsilon being 8.85*10 (permittivity constant)

well any help in solving this problem is very grateful
 
Last edited:
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You can solve for the enclosed charge, since the elementary charge of an electron (-1.60*10^-19 C) and the number of electrons is known. Just use Q/epsilon to solve for flux.
 
oh wow that was easy, i always see things harder then they are, thanks terbum
 

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