- #1
Ssnow
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- Is this a complicated expression for the ordinary Gauss theorem for the flux of the electric field ?
Hi to all!
The ordinary Gauss theorem states that ##\Phi\left(\vec{E}\right)\,=\, \frac{\sum_{i=1}^{n}q_{i}}{\varepsilon_{0}}## where ##\sum_{i=1}^{n}q_{i}## is the sum of all charges internal of a closed surface and ##\varepsilon_{0}## is the dielectric constant in the empty. Now I ask to the PF if this formula:
##\Phi\left(\vec{E}\right)\,=\, sign{\left(\sum_{i=1}^{n}q_{i}\right)}\cdot \left(\int_{-\infty}^{+\infty}e^{-\frac{\pi\varepsilon_{0}}{\left|\sum_{i=1}^{n}q_{i}\right|}y^2}dy\right)^2##
is equivalent to the previous and if it is mathematically correct.
Thank you!
Ssnow
The ordinary Gauss theorem states that ##\Phi\left(\vec{E}\right)\,=\, \frac{\sum_{i=1}^{n}q_{i}}{\varepsilon_{0}}## where ##\sum_{i=1}^{n}q_{i}## is the sum of all charges internal of a closed surface and ##\varepsilon_{0}## is the dielectric constant in the empty. Now I ask to the PF if this formula:
##\Phi\left(\vec{E}\right)\,=\, sign{\left(\sum_{i=1}^{n}q_{i}\right)}\cdot \left(\int_{-\infty}^{+\infty}e^{-\frac{\pi\varepsilon_{0}}{\left|\sum_{i=1}^{n}q_{i}\right|}y^2}dy\right)^2##
is equivalent to the previous and if it is mathematically correct.
Thank you!
Ssnow