I have seen this answer on why gauss's law can't be used to find the electric field due to a finite wire however, I'm unable to understand why translational symmetry is lost if we're only concerned with the field along the axis perpendicular to the wire's midpoint.
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I understand we require area to exploit guass's law basically, we could consider an infinitesimal area around the center , . – math and physics forever Sep 16 '22 at 12:21
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Short answer: in the finite wire case, there's an extra electric flux term due in the y-direction.
If you use Gauss's law,
$$\int E dS = Q/\epsilon_0$$
The integral term on the LHS involves the total electric flux contained within that imaginary cylinder.
In the case of the infinite wire, there was no flux in the y-direction since the wire runs continuous and we excluded that in our calculations.
However, in the finite wire case, we must include the flux term due to the electric field lines running in the y-direction. This is why calculating the electric field for a finite wire is more complex.
Karthik
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