why we don't use gauss theorem for open surface?
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Pick any electric field, even that due to just a single point charge. Draw the field lines around it, and draw the associated directions.
It is completely possible that if you select any surface that you select a surface with ALL the field lines pointing in the same direction. There is NO WAY that all of their field lines can add up to zero electric flux. You can't add exclusively positive numbers and get zero.
You can define electric flux for any surface, but there is no guarantee that it equals zero, unless that surface is a CLOSED surface.
Thats why we cant use this law on an open surface.
It is completely possible that if you select any surface that you select a surface with ALL the field lines pointing in the same direction. There is NO WAY that all of their field lines can add up to zero electric flux. You can't add exclusively positive numbers and get zero.
You can define electric flux for any surface, but there is no guarantee that it equals zero, unless that surface is a CLOSED surface.
Thats why we cant use this law on an open surface.
Answered by
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because reason is
fi net is = qinside/permitivity constant
if me take open surface qinside will be infinte because its contain all the charges present in whole universe because it is open surface to net fi me can is infinte. That'sby me use gauss theorem in close surface
fi net is = qinside/permitivity constant
if me take open surface qinside will be infinte because its contain all the charges present in whole universe because it is open surface to net fi me can is infinte. That'sby me use gauss theorem in close surface
samridhi17:
little bit. but thank you.
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