Friday, March 9, 2018

electrostatics - Why do outside charges do not contribute to net flux of a Gaussian Surface?


I don't quite understand why external charges can be ignored when calculating the net flux of a Gaussian surface. I understand that $\nabla \cdot \vec{E}$ of any point charge equals $0$ and I can reason using equations, but I can't find an intuitive physical understanding. Most arguments I have heard mention that all electric field lines that enter a Gaussian surface must then leave it, and so an external charge has no effect on net flux. But doesn't the flux also depend on the magnitude of the field?


For instance, Let's say I had a particle next to a Gaussian sphere and I look at the electric field line which pierces the sphere at its closest point. Wouldn't the field vector's magnitude be greater when it enters the sphere compared to when it exits because it is farther away when it leaves? And by the equation for flux,




$$\int \vec{E} \cdot \mathrm{d}\vec{A} = \int E \cos (\theta) \ \mathrm{d}A$$



which depends on $E$, wouldn't this have an affect on the net flux?


I'm not sure where my misunderstanding of flux is, but I know that I am clearly missing something huge. Perhaps is it that I have to consider all electric field lines and not just a single one? Or am I incorrectly assuming something about the relationship between the magnitude of the field and the flux through the surface?



Answer



If a charge is kept near a sphere, the charge will not affect the flux of the sphere because flux is dependent on magnitude of electric field and area it pass through. So when the field enters the near end of the sphere the magnitude of electric field is high and the surface pass through is low but when the filed comes out the magnitude of electric field is low but the area it pass through is high. Hence, it compensate and don't affect the flux of the sphere.


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