22 Replies
@Gyro Gearloose
in case snip aint clear
Note for OP
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why dont we do phi(out) + phi(in)
which book?
Area vector is outward for closed surfaces by convention
so for surface on 3a,0 flux is in +x dirn and for surface on 2a,0 flux is in -x dirn ?
Mhm
Well
Flux isn't in any direction
The area vector is
And thus the sign of the dot product is different
but dirn of phi(in) is diff in book
flux is defined as "number of field lines passing through the body"
thinking realistically at x=2a some field lines are going forward and at x=3a some more field lines are going forward so they add for that surface?
That definition is wrong.
The only definition of electric flux is the surface integral of the electric field over the area
Its not per say the number. A simple example is a field line of E and -E in the same direction. "The number of field lines" is 2 but flux is zero.
,rotate
can i interpret this scenario as in the pic above
on the right hand side there are much more field lines than left hand side (dont go by number )
through the lhs some field lines (say set A )
through the rhs field line of set A pass in addition to "new" field lines (say set B) produced due to some external reasons
now , since lines of set A come in and go out , they simply pass ( as in the case of charge q outside gaussian surface ) they dont produce any flux
the unique field lines of set B do produce the flux ( as in the case of charge q inside the gaussian surface )
acc to given relation on lhs we get setA and on rhs we get setA + setB . no obtain net flux we simply do setA + setB - setA i.e flux(rhs)-flux(lhs)= flux(out)-flux (in)
@SirLancelotDuLac is it correct?
from what my sir told me
we dont do phi(Out) + phi(in) because
the flux due to outside charge will enter and leave the surface
and the flux due to charge inside will leave
so to find out flux due to inside charge we subtract them
(im not 100% sure so if im wrong im sorry)
thats kinda what i wanted to express but with more detail and ease of understanding
H C VERMA
YouTube
Answer to a Students question on Gauss Law
Is the electric field appearing in Gauss's; law due to only the charges inside the closed surface or all the charges, inside and outside.
Interestingly both the options are correct!
cuz the field is pushing into the cube
+solved @chillout_ag
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